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NCEES-PE-Civil-WRE : NCEES PE Civil Engineering: Water Resources and Environmental Exam

NCEES NCEES-PE-Civil-WRE Questions & Answers

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NCEES PE Civil: Water Resources and Environmental


NCEES - PE Civil Engineering - Water Resources and Environmental 2024


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Question 386:


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/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

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?


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 µg/L to 100 µg/L over a distance of 50 m, what is the attenuation factor?


  1. 0.1

  2. 0.5

  3. 0.7

  4. 10.0


Answer: D


Explanation: The attenuation factor is calculated as:


C1 1000 μg/L

Attenuation Factor = = = 10

C2 100 μg/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,


Percentage Change =


8 − 5

8


× 100 = 37.5%

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 m³ and the total phosphorus concentration is 0.2 mg/L, what is the total phosphorus load in kilograms?


  1. 0.2 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/m³:


    Concentration = 0.2 mg/L = 0.0002 kg/m3


    Thus,


    Load = 0.0002 kg/m3 × 1, 000, 000 m3 = 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

    2 kg


aquifer has a hydraulic conductivity of 10 m/day, what is the estimated specific yield of the aquifer?


  1. 0.01

  2. 0.05

  3. 0.1

  4. 0.15

Answer: B


Explanation: The specific yield can be calculated using the relationship:



Specific Yield =

Drawdown

×

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 m³/yr

B. 1500 m³/yr

C. 1600 m³/yr

D. 1700 m³/yr


Answer: A


Explanation: The sustainable yield can be estimated using:

Time

1

Hydraulic Conductivity



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



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?

  1. 2 m

  2. 3 m

  3. 4 m

  4. 5 m



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 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?


A. 1288 kg

B. 1576 kg

C. 1036 kg

D. 1296 kg

Answer: D



Answer: C


Explanation: The mass of oxygen lost can be calculated as:


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



Reduction =


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

2200


× 100 ≈ 31.82%


Question 396:

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

Answer: D


Explanation: The total phosphorus load can be calculated as:


Load = Concentration × Volume


Convert concentration to kg/m³:


Load = 0.15 mg/L × 500, 000 m3 = 75 kg


Question 397:

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?


A. 0.15 mg

B. 0.25 mg

C. 0.50 mg

D. 1.50 mg

D. 75.0 kg



Answer: D


Explanation: The total mass can be calculated as:


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


Question 398:

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

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



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?


  1. 15.33 L/s/m

  2. 26.67 L/s/m

  3. 10.00 L/s/m

  4. 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 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


Question 401:


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 × Tim


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

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


e


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 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



Explanation: The nitrogen load can be calculated as:


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

Answer: C


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