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FE
environmental
practice exam
Ningdong Zhang (23-018-17) jessezhangningdong@outlook.com
Ningdong Zhang (23-018-17) jessezhangningdong@outlook.com
Copyright © 2020 by NCEES. All rights reserved.
All NCEES sample questions and solutions are copyrighted under the laws of the United States. No part of this publication may
be reproduced, stored in a retrieval system, or transmitted in any form or by any means without the prior written permission of
NCEES. Requests for permissions should be addressed in writing to permissions@ncees.org.
ISBN 978-1-932613-98-8
Printed in the United States of America
1st printing January 2020
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Ningdong Zhang (23-018-17) jessezhangningdong@outlook.com
CONTENTS
Introduction to NCEES Exams ................................................................ 1
About NCEES
Exam Format
Examinee Guide
Scoring and reporting
Updates on exam content and procedures
Exam Specifications .......................................................................... 3
Practice Exam .................................................................................... 7
Solutions .......................................................................................... 51
iii
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Ningdong Zhang (23-018-17) jessezhangningdong@outlook.com
About NCEES
NCEES is a nonprofit organization made up of the U.S. engineering and surveying licensing boards in
all 50 states, U.S. territories, and the District of Columbia. We develop and score the exams used for
engineering and surveying licensure in the United States. NCEES also promotes professional mobility
through its services for licensees and its member boards.
Engineering licensure in the United States is regulated by licensing boards in each state and territory.
These boards set and maintain the standards that protect the public they serve. As a result, licensing
requirements and procedures vary by jurisdiction, so stay in touch with your board (ncees.org/licensingboards).
Exam Format
The FE exam contains 110 questions and is administered year-round via computer at approved Pearson
VUE test centers. A 6-hour appointment time includes a tutorial, the exam, and a break. You’ll have
5 hours and 20 minutes to complete the actual exam.
In addition to traditional multiple-choice questions with one correct answer, the FE exam uses common
alternative item types such as
•
•
•
•
Multiple correct options—allows multiple choices to be correct
Point and click—requires examinees to click on part of a graphic to answer
Drag and drop—requires examinees to click on and drag items to match, sort, rank, or label
Fill in the blank—provides a space for examinees to enter a response to the question
To familiarize yourself with the format, style, and navigation of a computer-based exam, view the demo
on ncees.org/ExamPrep.
Examinee Guide
The NCEES Examinee Guide is the official guide to policies and procedures for all NCEES exams.
During exam registration and again on exam day, examinees must agree to abide by the conditions in
the Examinee Guide, which includes the CBT Examinee Rules and Agreement. You can download the
Examinee Guide at ncees.org/exams. It is your responsibility to make sure you have the current version.
Scoring and reporting
Exam results for computer-based exams are typically available 7–10 days after you take the exam. You
will receive an email notification from NCEES with instructions to view your results in your
MyNCEES account. All results are reported as pass or fail.
Updates on exam content and procedures
Visit us at ncees.org/exams for updates on everything exam-related, including specifications, exam-day
policies, scoring, and corrections to published exam preparation materials. This is also where you will
register for the exam and find additional steps you should follow in your state to be approved for the
exam.
1
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EXAM SPECIFICATIONS
•
3
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Fundamentals of Engineering (FE)
ENVIRONMENTAL CBT Exam Specifications
Effective Beginning with the July 2020 Examinations
•
The FE exam is a computer-based test (CBT). It is closed book with an electronic reference.
•
Examinees have 6 hours to complete the exam, which contains 110 questions. The 6-hour time also includes a
tutorial and an optional scheduled break.
•
The FE exam uses both the International System of Units (SI) and the U.S. Customary System (USCS).
Knowledge
Number of Questions
1.
Mathematics
A. Analytic geometry and trigonometry
B. Algebraic equations and roots
C. Calculus (e.g., differential, integral, differential equations)
D. Numerical methods (e.g., numerical integration, approximations, precision
limits, error propagation)
5–8
2.
Probability and Statistics
A. Measures of central tendencies and dispersions (e.g., mean, mode,
standard deviation)
B. Probability distributions (e.g., discrete, continuous, normal, binomial)
C. Estimation for a single mean (e.g., point, confidence intervals)
D. Regression (linear, multiple), curve fitting, and goodness of fit (e.g.,
correlation coefficient, least squares)
E. Hypothesis testing (e.g., t-test, outlier testing, analysis of the variance)
4–6
3.
Ethics and Professional Practice
A. Codes of ethics (e.g., professional and technical societies, ethical and
legal considerations)
B. Public health, safety, and welfare (e.g., public protection issues, licensing
boards, professional liability)
C. Compliance with codes, standards, and regulations (e.g., CWA, CAA, RCRA,
CERCLA, SDWA, NEPA, OSHA)
D. Engineer’s role in society (e.g., sustainability, resiliency, long-term viability)
5–8
4.
Engineering Economics
A. Time value of money (e.g., equivalence, present worth, equivalent annual
worth, future worth, rate of return, annuities)
B. Cost types and breakdowns (e.g., fixed, variable, direct and indirect labor,
incremental, average, sunk, O&M)
C. Economic analyses (e.g., benefit-cost, breakeven, minimum cost,
overhead, life cycle)
D. Project selection (e.g., comparison of projects with unequal lives,
lease/buy/make, depreciation, discounted cash flow)
5–8
4
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5.
Fundamental Principles
A. Population projections and demand calculations (e.g., water, wastewater,
solid waste, energy)
B. Reactors
C. Materials science (e.g., properties, corrosion, compatibility, stress strain)
7–11
6.
Environmental Chemistry
A. Stoichiometry and chemical reactions (e.g., equilibrium, acid-base,
oxidation-reduction, precipitation, pC-pH)
B. Kinetics (e.g., chemical conversion, growth and decay)
C. Organic chemistry (e.g., nomenclature, functional group reactions)
D. Multimedia equilibrium partitioning (e.g., Henry’s law, octanol
partitioning coefficient)
7–11
7.
Health Hazards and Risk Assessment
A. Dose-response toxicity (e.g., carcinogen, noncarcinogen)
B. Exposure routes and pathways
C. Occupational health (e.g., PPE, noise pollution, safety screening)
4–6
8.
Fluid Mechanics and Hydraulics
A. Fluid statics (e.g., pressure, force analysis)
B. Closed conduits (e.g., Darcy-Weisbach, Hazen-Williams, Moody)
C. Open channel (e.g., Manning, supercritical/subcritical, culverts,
hydraulic elements)
D. Pumps (e.g., power, operating point, parallel, series)
E. Flow measurement (e.g., weirs, orifices, flumes)
F. Blowers (e.g., power, inlet/outlet pressure, efficiency, operating point,
parallel, series)
G. Fluid dynamics (e.g., Bernoulli, laminar flow, turbulent flow,
continuity equation)
H. Steady and unsteady flow
12–18
9.
Thermodynamics
A. Thermodynamic laws (e.g., first law, second law)
B. Energy, heat, and work (e.g., efficiencies, coefficient of performance,
energy cycles, energy conversion, conduction, convection, radiation)
C. Behavior of ideal gases
3–5
10.
Surface Water Resources and Hydrology
A. Runoff calculations (e.g., land use, land cover, time of concentration,
duration, intensity, frequency, runoff control, runoff management)
B. Water storage sizing (e.g., reservoir, detention and retention basins)
C. Routing (e.g., channel, reservoir)
D. Water quality and modeling (e.g., erosion, channel stability, stormwater
quality management, wetlands, Streeter-Phelps, eutrophication)
E. Water budget (e.g., evapotranspiration, precipitation, infiltration, soil
moisture, storage)
9–14
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11.
Groundwater, Soils, and Sediments
A. Basic hydrogeology (e.g., aquifer properties, soil characteristics, subsurface)
B. Groundwater flow (e.g., Darcy’s law, specific capacity, velocity, gradient,
transport mechanisms)
C. Drawdown (e.g., Dupuit, Jacob, Theis, Thiem)
D. Remediation of soil, sediment, and/or groundwater (e.g., recovery,
ex-situ/in-situ treatment)
8–12
12.
Water and Wastewater
A. Water and wastewater characteristics (e.g., physical, chemical,
biological, nutrients)
B. Mass balance and loading rates (e.g., removal efficiencies)
C. Physical processes (e.g., sedimentation/clarification, filtration,
adsorption, membrane, flocculation, headworks, flow equalization, air
stripping, activated carbon)
D. Chemical processes (e.g., disinfection, ion exchange, softening,
coagulation, precipitation)
E. Biological processes (e.g., activated sludge, fixed film, lagoons,
phytoremediation, aerobic, anaerobic, anoxic)
F. Sludge treatment and handling (e.g., land application, digestion,
sludge dewatering, composting)
G. Water conservation and reuse
12–18
13.
Air Quality and Control
A. Ambient and indoor air quality (e.g., criteria, toxic and hazardous air
pollutants)
B. Mass and energy balances (e.g., STP basis, loading rates, heating values)
C. Emissions (e.g., factors, rates)
D. Atmospheric modeling and meteorology (e.g., stability classes, dispersion
modeling, lapse rates)
E. Gas treatment technologies (e.g., biofiltration, scrubbers, adsorbers,
incineration, catalytic reducers)
F. Particle treatment technologies (e.g., baghouses, cyclones,
electrostatic precipitators)
G. Indoor air quality modeling and controls (e.g., air exchanges, steadyand nonsteady-state reactor model)
8–12
14.
Solid and Hazardous Waste
A. Mass and energy balances
B. Solid waste management (e.g., collection, transportation, storage,
composting, recycling, waste to energy)
C. Solid waste disposal (e.g., landfills, leachate and gas collection)
D. Hazardous waste compatibility
E. Site characterization (e.g., sampling, monitoring, remedial investigation)
F. Hazardous and radioactive waste treatment and disposal (e.g., physical,
chemical, thermal, biological)
7–11
15.
Energy and Environment
A. Energy sources concepts (e.g., conventional and alternative)
B. Environmental impact of energy sources and production (e.g., greenhouse
gas production, carbon footprint, thermal, water needs)
4–6
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PRACTICE EXAM
•
•
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Ningdong Zhang (23-018-17) jessezhangningdong@outlook.com
FE ENVIRONMENTAL PRACTICE EXAM
1.
The equation of a sphere with center at (0, 1, −2) and a radius of 9 is:
o A.
o B.
o C.
o D.
2.
The roots of F =
o A.
o B.
o C.
o D.
3.
x 3 + 6 x 2 + 11x + 6
are most nearly:
x +1
–1, –2, –3
2, –3
–2, –3
2, 3
The solution to
o A.
o B.
o C.
o D.
4.
x2 + (y − 1)2 + (z + 2)2 = 81
x2 + (y + 1)2 + (z − 2)2 = 81
(x + 1)2 + (y + 1)2 + (z + 2)2 = 81
(x + 1)2 + (y + 1)2 + (z + 2)2 = 9
dy
+ 2y = 0 where y(0) = 3 is most nearly:
dt
y = 3e–2t
y = e–2t
y = −2
y = 3e–3t
If a = 3.5 ± 0.2, b = 2.0 ± 0.1, and c = 5.2 ± 0.3, and given the equation X = ab2(c)0.5, the value of
X is most nearly:
o A.
o B.
o C.
o D.
31.9 ± 0.1
31.9 ± 0.6
31.9 ± 5.3
31.9 ± 5.9
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FE ENVIRONMENTAL PRACTICE EXAM
5.
Given the function f(x) = 1/(x – 2), the linear approximation of f(x) around f(x) = –1 is most nearly:
o A.
o B.
o C.
o D.
6.
A series of measurements gives values of 11, 11, 11, 11, 12, 13, 13, 14, for which the arithmetic
mean is 12. The population standard deviation is most nearly:
o A.
o B.
o C.
o D.
7.
y = –1/9(x) – 4/9
y = –4/9(x) – 1/9
y = –1/3(x) + 4/9
y = –1/9(x) + 1/3
1.42
1.25
1.19
1.12
Two boxes each contain two blue blocks, one white block, and one green block. One block is
drawn from each box. The probability of drawing a white block from the first box and a blue block
from the second box is most nearly:
o A.
o B.
o C.
o D.
0.75
0.5
0.125
0.0625
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FE ENVIRONMENTAL PRACTICE EXAM
8.
Consider the following data:
Σ
x
2
3
5
8
9
11
13
15
66
y
0.3
2.5
2.1
3.6
3.1
6.5
5.5
6.0
29.6
xy
0.6
7.5
10.5
28.8
27.9
71.5
71.5
90.0
The least-squares fit linear equation is most nearly:
o A.
o B.
o C.
o D.
9.
y = 0.24 + 0.42x
y = 8.25 + 0.24x
y = 3.7 + 8.25x
y = 3.7 – 0.42x
Chlorine dosage into a wastewater treatment disinfection process must have a mean of 10 mg/L.
If testing leads the operator to believe that the dosages are too high or too small, the equipment
will be shut down and adjusted. The null hypothesis for not shutting down the equipment is:
o A.
o B.
o C.
o D.
Ho: μ = 10
Ho: μ ≠ 10
Ho: μ ≥ 10
Ho: μ < 10
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FE ENVIRONMENTAL PRACTICE EXAM
10.
11.
According to the Model Rules, Section 240.15, Rules of Professional Conduct, licensed
professional engineers are obligated to:
o A.
ensure that design documents and surveys are reviewed by a panel of licensed engineers
prior to affixing a seal of approval
o B.
express public opinions under the direction of an employer or client regardless of
knowledge of subject matter
o C.
practice by performing services only in the areas of their competence and in accordance
with the current standards of technical competence
o D.
do whatever is necessary to secure work, including offering, directly or indirectly,
services or other compensation in exchange for contracts
An engineer testifying as an expert witness in a product liability case should:
o A.
o B.
o C.
o D.
12.
answer as briefly as possible only those questions posed by the attorneys
provide a complete and objective analysis within his or her area of competence
provide an evaluation of the character of the defendant
provide information on the professional background of the defendant
In the context of hazardous waste regulations, joint and several liability means:
o A.
o B.
o C.
o D.
all responsible parties share cleanup costs equally
all responsible parties pay cleanup costs in proportion to their individual waste quantities
all responsible parties pay cleanup costs in proportion to the assessed value of their
individual properties
any individual responsible party may be required to pay the entire cost of cleanup
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FE ENVIRONMENTAL PRACTICE EXAM
13.
Polychlorinated biphenyls (PCBs) are primarily regulated under which of the following programs:
o A.
o B.
o C.
o D.
14.
Resource Conservation and Recovery Act (RCRA)
Federal Insecticide, Fungicide, and Rodenticide Act (FIFRA)
Comprehensive Environmental Response, Compensation, and Liability Act (CERCLA)
Toxic Substance Control Act (TSCA)
The Pollution Prevention Act of 1990 was passed to encourage pollution prevention in the United
States and established an environmental prevention hierarchy. Place the pollution prevention
methods in the hierarchy, from most preferred to least preferred.
Pollution Prevention Methods
Most Preferred
Least Preferred
15.
__________
Reuse/Recycle
__________
Disposal
__________
Treatment
__________
Source reduction
A tractor costs $7,500. After 10 years it has a salvage value of $5,000. Maintenance costs are
$500 per year. If the interest rate is 10%, the equivalent uniform annual cost is most nearly:
o A.
o B.
o C.
o D.
$500
$750
$1,400
$2,000
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FE ENVIRONMENTAL PRACTICE EXAM
16.
The relationship between the purchase price (P) per unit, in $, and consumer demand (D), in units
sold, for a certain component is defined by the equation P = 220 – 0.025D.
The fixed costs for the production of this unit are $75,000 per month, and the variable cost is $82
per unit. If 1,200 units per month are produced and sold to consumers, the resulting monthly profit
is $__________.
Enter your response in the blank.
17.
A company can manufacture a product using hand tools. Tools will cost $1,000, and the
manufacturing cost per unit will be $1.50. As an alternative, an automated system will cost $15,000
with a manufacturing cost per unit of $0.50. The anticipated annual volume is 5,000 units. If
interest is neglected, the breakeven point (years) is most nearly:
o A.
o B.
o C.
o D.
2.8
3.6
15.0
never
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FE ENVIRONMENTAL PRACTICE EXAM
18.
A company requires a certain type of equipment for 6 years. Two mutually exclusive options are
available for purchase. Equipment A has a 6-year life cycle and an initial cost of $330,000, and it
will generate revenue of $71,000 per year while in use. Equipment B has a 4-year life cycle and
an initial cost of $215,000, and it will generate revenue of $64,000 per year while in use. Leased
equipment can be used in years 5 and 6, at a cost of $63,000 per year, to generate revenue of
$64,000 per year.
For an interest rate of 6%, the present worth of the less profitable alternative is most nearly:
o A.
o B.
o C.
o D.
$6,770
$8,220
$19,130
$43,000
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FE ENVIRONMENTAL PRACTICE EXAM
19.
You must choose between four pieces of comparable equipment based on the cash flows given
below. All four pieces have a life of 8 yr.
First cost
A
$25,000
Equipment
B
C
$35,000
$20,000
D
$40,000
Annual costs
$8,000
$6,000
$9,000
$5,000
Salvage value
$2,500
$3,500
$2,000
$4,000
Parameter
The discount rate is 12%. Ignore taxes. The two most preferable projects and the difference
between their present worth values are most nearly:
o A.
o B.
o C.
o D.
20.
A and C, $170
B and D, $170
A and C, $234
B and D, $234
A community has experienced the following growth in population and water consumption:
Year
2005
2015
Population Water Consumption
(MGD)
45,000
6.30
61,000
9.45
Assume that the annual percentage rate of population growth remains equal over time, but the per
capita water consumption rate stabilizes at 160 gallons per day. The projected water demand
(MGD) in 2025 is most nearly:
o A.
o B.
o C.
o D.
12.3
12.6
13.2
14.2
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FE ENVIRONMENTAL PRACTICE EXAM
21.
A village has a current population of 2,000 people. The village census 10 years ago showed a
population of 1,820 people. An anticipated project has a design life of 30 years. Assuming the
village maintains the same linear growth rate from the past 10 years, the population at the end of
the project's design is projected to most nearly be:
o A.
o B.
o C.
o D.
22.
A city is planning to increase the size of its reservoir due to projected population growth. The city
has recorded populations of 150,000 in January of 1999 and 225,000 in January of 2009. Based on
exponential growth, the population the city should plan for in Year 2025 is most nearly:
o A.
o B.
o C.
o D.
23.
2,650
2,540
2,360
2,000
345,000
375,000
430,000
450,000
Oxidation of iron in a groundwater supply is described by first-order kinetics with a rate constant
of 5/hr. The untreated iron concentration is 1.8 mg/L, and 90% removal is desired. The necessary
hydraulic residence time (hr) for a complete mix reactor is most nearly:
o A.
o B.
o C.
o D.
0.02
0.18
0.46
1.80
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FE ENVIRONMENTAL PRACTICE EXAM
24.
A disinfection process for potable water treatment is described by first-order kinetics with a rate
constant of 0.1 per minute. An ideal plug flow reactor is used with a hydraulic residence time of
1 hr. The expected efficiency of the system is most nearly:
o A.
o B.
o C.
o D.
25.
The pressure gauge in an air cylinder reads 1,680 kPa. The cylinder is constructed of a 12-mm
rolled-steel plate with an internal diameter of 700 mm. The tangential stress (MPa) inside the tank
is most nearly:
o A.
o B.
o C.
o D.
26.
90.0%
99.0%
99.9%
99.99%
25
50
77
100
If an aluminum crimp connector is used to connect a copper wire to a battery, what would you
expect to happen?
o A.
o B.
o C.
o D.
Only the copper wire will corrode.
Only the aluminum connector will corrode.
Both will corrode.
Nothing
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FE ENVIRONMENTAL PRACTICE EXAM
27.
Consider the following stoichiometric chemical equation:
4A + B ↔ 2C + 2D
Which of the following is the correct chemical equilibrium constant equation?
28.
o A.
4
A] [ B ]
[
K=
[C ]2 [ D ]2
o B.
K=
o C.
2
2
C ] [ D]
[
K=
[ A]4 [ B ]
o D.
K=
[ 4 A][ B ]
[ 2C ][ 2 D ]
[ 2C ][ 2 D ]
[ 4 A][ B ]
A municipal wastewater treatment facility effluent contains 8 mg/L of BOD5. The reaction rate
constant for BOD exertion (base e) is known to be 0.10 day–1 for this effluent. The ultimate
BOD (mg/L) of the effluent is most nearly:
o A.
o B.
o C.
o D.
3
12
20
30
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FE ENVIRONMENTAL PRACTICE EXAM
29.
What kind of chemical bonding is represented by the molecules of cyclohexane, cyclohexene,
and methanol?
o A.
o B.
o C.
o D.
30.
The concentration of TCE in the octanol phase of a two-phase octanol/water sample is 15 µg/L,
and the log K OWTCE = 2.3 . The concentration of TCE (µg/L) in the aqueous phase is most nearly:
o A.
o B.
o C.
o D.
31.
Covalent
Ionic
Hydrogen
Metallic
0.1
0.2
6.5
13.3
How is a solution with a pH of 2 prepared using 6M HCl?
o A.
o B.
o C.
o D.
Dilute 1.67 mL of HCl to 1 L, using water.
Dilute 6 mL of HCl to 1 L, using water.
Dilute 16.7 mL of HCl to 1 L, using water.
Dilute 60 mL of HCl to 1 L, using water.
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FE ENVIRONMENTAL PRACTICE EXAM
32.
The activity of a radionuclide was measured at 1,200 Bq. Thirty-two days later, the activity was
measured at 150 Bq. The half-life (days) of the radionuclide is most nearly:
o A.
o B.
o C.
o D.
33.
A typical example of an arene is:
o A.
o B.
o C.
o D.
34.
acetylene
benzene
methylamine
dimethyl
A carcinogen is in drinking water at a concentration of 0.01 mg/L. The cancer risk of 30 years of
adult (75 kg) exposure, given a cancer slope factor of 0.80 [mg/(kg·day)]–1, is most nearly:
o A.
o B.
o C.
o D.
35.
4
5
11
16
2.4 × 10–1
1.2 × 10–4
9.8 × 10–5
4.1 × 10–6
In general, dermal contact is a route of exposure for each of the following environmental
media except:
o A.
o B.
o C.
o D.
groundwater
surface water
soil
food
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FE ENVIRONMENTAL PRACTICE EXAM
36.
A manufacturing plant wants to install a new piece of machinery outdoors. The nearest resident
would be located 500 m from the equipment. The sound pressure level was measured to be
100 dBA at a distance of 5 m from the equipment. The resulting sound pressure level (dBA) at a
distance of 500 m from the machine is most nearly:
o A.
o B.
o C.
o D.
37.
10
40
60
80
A worker is exposed to a toxic (noncarcinogenic) compound for a duration of 40 hr/week. The
concentration of the chemical in the workplace air is 40 μg/m3, and the worker is assumed to inhale
at a rate of 0.9 m3/hr. The individual has a body mass of 70 kg. The chronic daily intake
[mg/(kg·day)] during the period of occupational exposure is most nearly:
o A.
o B.
o C.
o D.
0.001
0.003
0.012
0.088
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FE ENVIRONMENTAL PRACTICE EXAM
38.
The rectangular homogeneous gate shown below is 3.00 m high × 1.00 m wide and has a
frictionless hinge at the bottom. If the fluid on the left side of the gate has a density of 1,600 kg/m3,
the magnitude of the force F (kN) required to keep the gate closed is most nearly:
F
FLUID
3.00 m
FRICTIONLESS
HINGE
o A.
o B.
o C.
o D.
39.
0
22
24
220
Two open tanks are connected by a single pipe. The water surface elevations of the upstream and
downstream tank are 103.00 ft and 101.00 ft, respectively. The pipe is 60 ft long and 6 inches in
diameter. The Darcy friction factor f is 0.020. The pipe is connected to the tanks with a sharp
entrance and a sharp exit. There are no other fittings. The expected flow rate in the pipe (cfs) is
most nearly:
o A.
o B.
o C.
o D.
1.13
1.44
1.71
5.75
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FE ENVIRONMENTAL PRACTICE EXAM
40.
A 10-ft-wide rectangular concrete channel is to be built to convey storm water runoff. The channel
grade is 0.2%. The Manning's roughness coefficient n is 0.015. If the depth of flow is 2 ft, the
discharge (cfs) is most nearly:
o A.
o B.
o C.
o D.
41.
110
150
250
500
Mark the point on the system curve that shows the head and flow when the two pumps, Pumps A
and Pump B, are working in parallel in the system.
200
PUMPS IN SERIES
PERFORMANCE B
PUMP CURVE
TOTAL HEAD (ft)
150
PERFORMANCE A
PUMP CURVE
100
SYSTEM CURVE
50
PUMPS IN PARALLEL
0
100
200
300
400
500
FLOW RATE (gpm)
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FE ENVIRONMENTAL PRACTICE EXAM
42.
The pitot tube shown below is placed at a point where the velocity is 2.0 m/s. The specific gravity
of the fluid is 2.0, and the upper portion of the manometer contains air. The reading h (m) on the
manometer is most nearly:
AIR
h
2.0 m/s
o A.
o B.
o C.
o D.
43.
A 200-hp blower operating at 81% efficiency is supplying air at a rate of 2 lb/sec. The inlet pressure
is 14.7 psia, and the inlet temperature is 112oF. The outlet pressure (psia) produced by the blower
is most nearly:
o A.
o B.
o C.
o D.
44.
20.0
10.0
0.40
0.20
61
50
21
16
The difference between the energy line and the hydraulic grade line for a steady incompressible
flow in a closed pipe under pressure is:
o A.
o B.
o C.
o D.
pressure head
elevation head
velocity head
headloss
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FE ENVIRONMENTAL PRACTICE EXAM
45.
A rapid mixing process must provide a minimum hydraulic residence time of 45 sec. The water
production rate is 0.30 MGD. Two identical tanks are provided in parallel. If the tanks are
cylindrical with a liquid depth equal to the radius, the minimum liquid depth (ft) is most nearly:
o A.
o B.
o C.
o D.
46.
1.50
1.82
2.19
2.92
A pump is used to deliver water from a lake to an elevated storage tank. The elevations of the
following components are known:
Lake water surface = 93.00 ft
Pump centerline = 99.00 ft
Ground surface at storage tank = 109.00 ft
Water surface in storage tank = 199.00 ft
The static head (ft) for this pumping application is most nearly:
o A.
o B.
o C.
o D.
6
16
100
106
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FE ENVIRONMENTAL PRACTICE EXAM
47.
A flow of 15.5 cfs enters the pipe system at A as shown, and exits at B and C. Pipe data are given
in the following table.
Assume the pipes are all at the same elevation. The head loss (ft) in Pipe CB is most nearly:
o A.
o B.
o C.
o D.
0.05
0.13
0.18
0.23
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FE ENVIRONMENTAL PRACTICE EXAM
48.
A sanitary sewer has the following characteristics:
Manning's roughness coefficient (n) = 0.014
Slope (s) = 0.5%
Diameter = 1 m
The discharge (m3/s) when the sewer is flowing full is most nearly:
o A.
o B.
o C.
o D.
49.
Water (density = 1,000 kg/m3) flows at a rate of 1.50 m3/s through a horizontal pipe having a crosssectional area of 0.25 m2. The flow rate is currently increasing at a rate of 0.50 m3/s per second.
For a pressure of 100,000 N/m2 at A, and neglecting pipe friction, the current pressure (N/m2)
directly downstream at B is most nearly:
o A.
o B.
o C.
o D.
50.
1.6
2.3
2.5
15.7
94,000
95,500
99,250
118,000
A system is designed to allow at least 100,000 Btu/min of heat to be transferred (Q) from the air
to the system as air flows through it. The system can be operated so that air (cp = 0.26 Btu/lb-F)
flowing at 500 lb/min will exit at 100°F. The minimum temperature (°F) the air can enter the
system is
.
Enter your response in the blank.
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FE ENVIRONMENTAL PRACTICE EXAM
51.
Hot air at 200°C flows across a 50°C surface. If the heat transfer coefficient is 72 W/(m2·°C), the
heat-transfer rate (W) over 2 m2 of the surface is most nearly:
o A.
o B.
o C.
o D.
52.
When the pressure of a constant mass of an ideal gas is doubled and the absolute temperature is
halved, the volume is:
o A.
o B.
o C.
o D.
53.
300
5,625
11,250
21,600
quadrupled
no change
halved
quartered
A 35-acre watershed (curve number = 73) receives 4.1 in. of precipitation. The runoff volume (ft3)
is most nearly:
o A.
o B.
o C.
o D.
51,000
144,000
203,000
380,000
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FE ENVIRONMENTAL PRACTICE EXAM
54.
A community of 35,000 people requires a reservoir to supplement the available surface water
during an annual dry season. The available surface water of the driest year is shown.
Month
Available
Surface Water
(MGD)
January
4.2
February
4.7
March
5.9
April
6.3
May
3.8
June
1.2
July
1.0
August
0.8
September
2.9
October
3.8
November
3.8
December
4.0
Assuming a demand of 80 gallons per capita per day (gpcd) and 30 days per month, the required
reservoir size (millions of gallons) is most nearly:
o A.
o B.
o C.
o D.
2.8
60
162
486
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FE ENVIRONMENTAL PRACTICE EXAM
55.
Stormwater runoff is routed through a pond having the following characteristics:
Water Depth
(ft)
0.0
1.0
2.0
3.0
4.0
5.0
Volume of Water in Pond
(ft3)
0
19,100
43,200
57,600
75,600
163,400
Outflow
(cfs)
0.0
3.0
5.0
7.0
9.0
12.0
During a certain storm event, runoff to the pond is 16 cfs during the first hour, 12 cfs during the
second hour, and 0 cfs during the third hour. Use the volume of water in the pond at the end of
hour n – 1 to determine the outflow during hour n. Assuming that the pond is empty when the
storm event begins, the water depth (ft) at the end of the third hour is most nearly:
o A
o B
o C
o D
56.
2.0
3.0
4.0
5.0
A stream receives a discharge from an industrial wastewater treatment plant. After mixing in the
stream, the ultimate BOD is 15 mg/L, the dissolved oxygen is 7 mg/L, and the temperature is 20°C.
At the stream temperature, the saturated dissolved oxygen concentration is 9.2 mg/L. The
deoxygenation rate constant (base e) and reaeration rate constant (base e) are 0.2 day–1 and
0.4 day–1, respectively. The minimum dissolved oxygen concentration (mg/L) in the stream
downstream from this discharge is most nearly:
o A.
o B.
o C.
o D.
0
4.4
4.8
5.0
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FE ENVIRONMENTAL PRACTICE EXAM
57.
A reservoir used for a 2.5-MGD municipal water supply contains an initial volume of 205 Mgal
of water. On an annual basis, there is an average of 3 ft of precipitation runoff and 16.2 in. of
evapotranspiration over the 2,000-acre drainage basin (including the reservoir surface area). There
is also 7.8 in. of seepage loss out of the 21-acre reservoir. The volume of water (millions of gallons)
at the end of the year is most nearly:
o A.
o B.
o C.
o D.
118
349
363
1,236
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FE ENVIRONMENTAL PRACTICE EXAM
58.
Intensity-duration rainfall curves are shown below. A 12-acre residential development has a runoff
coefficient C of 0.40. The time of concentration for runoff from the development is 1 hr. If the
rational method is used, the peak runoff (cfs) from the site for a 10-yr storm is most nearly:
QUANTITY OF STORMWATER
7
6
15
5
4
RAINFALL INTENSITY
(cm/hr)
3
5
2 0
15 5
1
50
40
60
Y
UE
(ye
ars
)
RAINFALL INTENSITY
(in./hr)
10
EQ
FR
1
0
5
NC
2
0
20
2
0
80 100 120 140 160 180 200
TIME (min)
o A.
o B.
o C.
o D.
8.6
11.0
13.0
21.6
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FE ENVIRONMENTAL PRACTICE EXAM
59.
A reservoir is to be designed for a water supply system that serves a population of 300,000 people,
with an average demand of 100 gal/day/capita and a required fire flow of 3,000 gal/min for a
duration of 8 hours. If the maximum daily demand peak factor is 1.5, the required volume (millions
of gallons) of storage in the reservoir is most nearly:
o A.
o B.
o C.
o D.
60.
A municipal wastewater discharge into a reservoir must be regulated to prevent eutrophication.
The discharge permit would likely include limits on which of the following parameters?
o A.
o B.
o C.
o D.
61.
30
45
46.4
67.5
BOD5, SS, Coliforms
BOD5, SS, Coliforms, TKN, Total P
BOD5, SS, Coliforms, Total N, Total P
BOD5, SS, Coliforms, Total N, Total P, Total K
In an area with a composite runoff coefficient of 0.65, the surface runoff flows toward a street
from the land on both sides. The watershed area extends to 100 ft on each side of the street
centerline. The street has curb-and-gutter, and both sides of the street have a curb inlet (or basin).
The capacity of the curb inlet to pick up runoff from the gutter is 10 cfs (any more than this will
just run past the opening). City policy is to design the street drainage system to accommodate a
6.8 in./hr rainfall. The distance (ft) between the inlets along the street should be most nearly:
o A.
o B.
o C.
o D.
980
640
490
230
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FE ENVIRONMENTAL PRACTICE EXAM
62.
Which of the following statements is true of facultative bacteria?
o A.
o B.
o C.
o D.
63.
Facultative bacteria require oxygen as an electron acceptor for metabolism.
Facultative bacteria cannot use oxygen as an electron acceptor for metabolism.
Facultative bacteria require nitrate as an electron acceptor for metabolism.
None of the above
A rapid sand filter has four cells with the following dimensions:
Length = 10 ft
Width = 10 ft
Depth = 12 ft
The water production rate is 2 MGD. If one cell of the filter is taken offline for backwash, the
hydraulic loading rate (gpm/ft2) for the filter is most nearly:
o A.
o B.
o C.
o D.
64.
3.5
4.6
6.2
7.7
A lime soda softening plant produces a municipal supply of 5 MGD. The plant includes two
clarifiers, each with a diameter of 65 ft and sidewater depth of 12 ft. The clarifier residence
time (hr) for parallel operation is most nearly:
o A.
o B.
o C.
o D.
0.35
1.43
2.86
3.19
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FE ENVIRONMENTAL PRACTICE EXAM
65.
The primary origin of trihalomethanes in potable water supplies is:
o A.
o B.
o C.
o D.
66.
contamination of groundwater with industrial pollutants
corrosion products in the water distribution system
by-products of chlorination from disinfection
contamination of surface water with agricultural runoff
A complete mix of an activated-sludge system without primary clarification is used for treatment
of a municipal wastewater with a flow rate of 12 MGD. The plant components include four
clarifiers with a diameter of 80 ft and a side-water depth of 14 ft. The aeration basin volume is
3 million gallons, the mixed liquor suspended solids concentration is 3,000 mg/L, and the solids
residence time is 5 days. The return activated-sludge flow rate is 10 MGD. The effluent suspended
solids concentration is negligible. Sludge is wasted from the aeration basin.
The return sludge suspended solids concentration (mg/L) is most nearly:
o A.
o B.
o C.
o D.
3,000
6,600
10,000
15,000
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FE ENVIRONMENTAL PRACTICE EXAM
67.
An aeration basin for an activated-sludge facility has the following characteristics:
Length = 90 ft
Width = 30 ft
Liquid depth = 12 ft
MLSS = 4,000 mg/L
The raw wastewater has the following characteristics:
Flow = 1.2 MGD
BOD5 = 220 mg/L
Suspended solids (SS) = 250 mg/L
A primary clarifier is provided that removes 25% of the BOD5 and 60% of the suspended solids.
The return activated-sludge flow rate is 0.8 MGD.
If the primary sludge solids content is 4% solids and the specific gravity is 1.0, the primary sludge
volume (ft3/day) is most nearly:
o A.
o B.
o C.
o D.
68.
600
1,000
4,500
60,000
Match the treatment technology that will best achieve the treatment objective for the potable reuse
of water. Use each technology only once.
Treatment Objective
Treatment Technology
A. Remove suspended solids
1. UV/H2O2
B. Reduce dissolved nutrients
2. Media filtration
C. Degrade trace organics
3. Biological mediation
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FE ENVIRONMENTAL PRACTICE EXAM
69.
Based on the following information, the annual CO2 emissions (metric tons) for a home are most
nearly:
Parameter
Annual electricity consumed in the home
Value
12,395 kWh
Average CO2 emission from electricity generation 900 lb CO2/MWh
Average losses of generated electricity due to
transmission and distribution
o A.
o B.
o C.
o D.
70.
8%
5.0
5.5
6.0
6.5
Three wastewater flows combine in a sewer, each having flows and BOD concentrations as
follows.
Source
Flow (L/day)
BOD (mg/L)
1
4.0 × 106
200
2
0.8 × 106
300
3
0.2 × 106
500
If infiltration (having zero BOD) is 10% of total flow, the resulting BOD (mg/L) is most nearly:
o A.
o B.
o C.
o D.
76
207
228
333
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FE ENVIRONMENTAL PRACTICE EXAM
71.
Place the unit operations and processes in the most likely sequence for a surface water treatment
facility.
Strategies
First
Last
72.
_______________________
Disinfection
_______________________
Sedimentation
_______________________
Flocculation
_______________________
Rapid Mix
_______________________
Filtration
A groundwater contains the following cations:
Sodium = 40 mg/L
Calcium = 70 mg/L
Magnesium = 30 mg/L
The total hardness (mg/L as CaCO3) is ____________.
Enter your response in the blank.
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FE ENVIRONMENTAL PRACTICE EXAM
73.
An aeration basin for an activated-sludge facility has the following characteristics:
Length = 90 ft
Width = 30 ft
Liquid depth = 12 ft
Mixed-liquor suspended solids (MLSS) = 4,000 mg/L
The 1.2-MGD flow of wastewater from the primary clarifier has the following characteristics:
BOD5 = 165 mg/L
Suspended solids (SS) = 100 mg/L
The aeration basin organic loading rate (F:M) (lb BOD5/lb SS-day) is most nearly:
o A.
o B.
o C.
o D.
74.
0.038
0.055
0.20
0.27
A radiant kerosene heater is used to heat a living space. When selecting a home alarm, you would
want to ensure that it monitors which indoor air pollutant?
o A.
o B.
o C.
o D.
Radon
Carbon monoxide
Carbon dioxide
Formaldehyde
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FE ENVIRONMENTAL PRACTICE EXAM
75.
An 8-in.-diameter duct is used to collect and vent emissions from a heated process tank. At the
duct inlet, the air temperature is 130°F and the mass flow rate of air is 75 lb/min. The gas
temperature is reduced to 110°F during travel to the duct outlet. The mass flow rate of air (lb/min)
leaving the duct is most nearly:
o A.
o B.
o C.
o D.
76.
Operation of a hydrocarbon combustion process with an increased air-to-fuel ratio from
stoichiometric to slightly above stoichiometric would be expected to increase emissions of:
o A.
o B.
o C.
o D.
77.
64
72
75
78
CO
VOC
PM10
NOx
The ratio of vapor pressure to the saturation pressure of water at a given temperature can be used
to calculate:
o A.
o B.
o C.
o D.
relative humidity
specific humidity
dew-point temperature
wet-bulb temperature
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FE ENVIRONMENTAL PRACTICE EXAM
78.
For coal-fired steam electric power plants, control of SOx emissions is most commonly achieved
by:
o A.
o B.
o C.
o D.
79.
lime scrubbing
catalytic conversion
electrostatic precipitation
carbon adsorption
A baghouse is considered an appropriate air pollution control device for which of the following
pollutant streams?
Select all that apply.
□ A.
□ B.
□ C.
□ D.
□ E.
□ F.
□ G.
80.
Carbon black
Carbon monoxide
Fly ash
Formaldehyde
Grains
Paint pigments
Radon
A manufacturing plant assembly room has the following dimensions: 50 m × 20 m × 10 m.
Evaporative solvent emissions were measured to equal 1.0 kg/hr. The production area is ventilated
at a rate of 80 air changes per hour. Assume no solvent is present in the outside air and that the
solvent concentration is uniform throughout the assembly room. The workplace air solvent
concentration (μg/m3) is most nearly:
o A.
o B.
o C.
o D.
1
100
1,250
100,000
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FE ENVIRONMENTAL PRACTICE EXAM
81.
Emissions of a solvent from an industrial source at ground level are 100 μg/s. If the wind speed is
3 m/s and the atmosphere has a neutral stability, the ground-level centerline concentration (μg/m3)
0.1 km downwind from the source is most nearly:
o A.
o B.
o C.
o D.
82.
0.3
10
33
300
Analysis of the municipal solid waste for a community with a population of 50,000 revealed the
following composition (mass basis):
Paper products = 35%
Yard wastes = 20%
Food wastes = 10%
Plastics = 9%
Metals = 8%
Wood = 5%
Glass = 5%
Other = 8%
Implementation of a curbside recycling program is estimated to achieve 40% recycle of paper
products, 20% recycle of metals, and 30% recycle of glass. Separate collection and composting of
yard wastes is estimated to reduce quantities by 80%. Implementation of the curbside recycling
and yard waste segregation programs would achieve a reduction in the mass of municipal solid
waste of most nearly:
o A.
o B.
o C.
o D.
17%
33%
50%
65%
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FE ENVIRONMENTAL PRACTICE EXAM
83.
A 10-acre surface area pond is located within a 640-acre watershed. The pond is in a tight clay soil
so seepage is negligible, and there are no plants around the pond margins. During the 30-day month
of June, the watershed receives 2.50 in. of rainfall. Average stream flows into and out of the pond
during the month are 0.450 cfs and 0.10 cfs, respectively. The pond elevation at the beginning of
the month is 512.9 ft, mean sea level (MSL). By the end of the month, the pond has risen to
514.8 ft, MSL. The amount of evaporation that takes place in inches of water from the surface of
the pond during the month is most nearly ________________________.
Enter your response in the blank.
84.
A community generates 50,000 lb/day of solid waste that is disposed of in a sanitary landfill. The
mass ratio of refuse to cover is 3 to 1. The in-place density of the fill (refuse plus cover) is
1,000 lb/yd3. The necessary volume (yd3) of fill (refuse plus cover) for a 10-year operation period
is most nearly:
o A.
o B.
o C.
o D.
730,000
243,000
183,000
61,000
85.
Which of the following would most likely result in an explosion if the materials were mixed?
o
o
o
o
A.
B.
C.
D.
An isocyanine and an amide
Hydrazine and an organic sulfide
A caustic and an ester
Halogenated organics, alkaline earth metal, and elemental metal
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FE ENVIRONMENTAL PRACTICE EXAM
86.
You are conducting a preliminary site investigation and must develop the soil sampling plan. Given
the following guidelines, the minimum number of samples required is:
Confidence level = 80%
Power = 90%
Standard deviation of sample = 0.1
Sample average = 0.4
Minimum detectable relative difference = 20%
o A.
o B.
o C.
o D.
87.
4
8
12
15
Mark the diamond which signifies that a material will ignite if moderately heated.
2
2
2
2
88.
A solid waste transfer bin vehicle has a load capacity of 16 tons. Refuse collection trucks have a
volume of 8 yd3 at a density of 0.25 ton/yd3. In order to fill each transfer bin vehicle to capacity,
the number of refuse collection trucks required is ________________.
Enter your response in the blank.
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FE ENVIRONMENTAL PRACTICE EXAM
89.
A saturated sandy soil has a porosity of 40%. The specific gravity of the sand is 2.65. The specific
weight (lb/ft3) of the saturated soil is most nearly:
o A.
o B.
o C.
o D.
90.
A constant head permeameter test has a soil specimen diameter of 4 cm and a hydraulic gradient
of 4. If the flow is constant at 25 cm3/day, the hydraulic conductivity (cm/day) is most nearly:
o A.
o B.
o C.
o D.
91.
99
124
136
165
0.01
0.50
1.56
2.00
Which equation would you use to calculate the flow rate of water drawn from a well?
o A.
o B.
o C.
o D.
Dupuit formula
Bernoulli equation
Rational equation
Hazen-Williams equation
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FE ENVIRONMENTAL PRACTICE EXAM
92.
An air stripper will be used for reducing Compound X from a contaminated groundwater
containing 275 μg/L to a level of 2 μg/L.
Dimensionless Henry's constant for X = 0.5
KLa for stripper media = 50 hr–1
Qair = 100 m3/s
Qwater = 1 m3/s
kmol
Molar loading of X = 10
s m 2
The required height (m) of the stripper packing is most nearly:
o A.
o B.
o C.
o D.
93.
A piezometer located in an unconfined aquifer is monitored on a monthly basis. The response time
associated with rainfall precipitation in an unconfined aquifer, in general, is considered to be:
o A.
o B.
o C.
o D.
94.
74
65
18
1
shorter than that of a confined aquifer
the same as that of a confined aquifer
longer than that of a confined aquifer
cannot be determined without installing additional piezometers
Two monitoring wells are constructed in an unconfined aquifer. The wells are separated by a
distance of 250 ft. The water surface elevations in the up-gradient and down-gradient wells are
101.00 ft and 100.85 ft, respectively. The aquifer hydraulic conductivity is 5 ft/day. The fluid
velocity (ft/day) in the aquifer is most nearly:
o A.
o B.
o C.
o D.
0.0006
0.003
0.75
5
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FE ENVIRONMENTAL PRACTICE EXAM
95.
A fully penetrating groundwater extraction well is used to capture a contaminant plume in an
unconfined aquifer with a hydraulic conductivity of 5 × 10–5 ft/sec. During steady pumping of
1 × 10–2 ft3/sec, a cone of depression is formed as shown. The diameter (in.) of the well is most
nearly:
30 ft
Q
20 ft
10 ft
o A.
o B.
o C.
o D.
0.27
0.54
3.2
6.5
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FE ENVIRONMENTAL PRACTICE EXAM
96.
The transmissivity of a 45-m-thick confined aquifer is 1.6 × 10–2 m2/s. Two groundwater
monitoring wells are placed 50 m apart in the direction of flow. The measured difference in
hydraulic head between the two wells is 37 cm. The discharge rate (m3/day) through a 15-m-wide
area of interest is most nearly:
WELL 1
WELL 2
GROUNDWATER
FLOW
15 m
o A.
o B.
o C.
o D.
97.
0.0018
150
6,900
7,700
The analysis of a 100-lb municipal refuse sample is provided.
Component
Food waste
Paper
Plastic
Wood
Weight
(lb)
20
40
10
30
Component Energy Value
(BTU/lb)
2,500
6,500
12,000
7,500
The energy value (BTU/lb) of the entire sample is _____________.
Enter your response in the blank.
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FE ENVIRONMENTAL PRACTICE EXAM
98.
An environmental impact study of a proposed wind turbine operation would consider the potential
for all of the following EXCEPT increased:
o A.
o B.
o C.
o D.
99.
NOx emissions
shadow flickering
bat collisions
noise pollution
A 1,000-MW steam electric power plant operates with a thermal efficiency of 33%. The plant
burns coal with an energy content of 11,500 Btu/lb. New source performance standards permit
emission of 0.60 lb NOx per million Btu of heat input. The maximum allowable rate of emission
of NOx (lb/day) is most nearly:
o A.
o B.
o C.
o D.
100.
6,200
49,600
74,400
148,800
The mass (kg) of pure oxygen theoretically required to completely burn 3 kg of benzene (C6H6) is
most nearly:
o A.
o B.
o C.
o D.
720
9.2
7.4
4.6
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SOLUTIONS
•
•
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FE ENVIRONMENTAL SOLUTIONS
Detailed solutions for each question begin on the next page.
1
A
26
B
51
D
76
D
2
C
27
C
52
D
77
A
3
A
28
C
53
C
78
A
4
D
29
A
54
C
79
A, C, E, F
5
A
30
A
55
A
80
C
6
D
31
A
56
C
81
A
7
C
32
C
57
C
82
B
8
A
33
B
58
A
83
4.2–5.2 in.
9
A
34
C
59
C
84
B
10
C
35
D
60
C
85
D
11
B
36
C
61
A
86
B
12
D
37
B
62
D
87
see solution
13
D
38
C
63
B
88
8
14
see solution
39
A
64
C
89
B
15
C
40
A
65
C
90
B
16
$54,600
41
see solution
66
B
91
A
17
A
42
D
67
A
92
B
18
B
43
B
68
see solution
93
A
19
B
44
C
69
B
94
B
20
C
45
A
70
B
95
D
21
B
46
D
71
see solution
96
B
22
C
47
B
72
297–300
97
6,400–6,600
23
D
48
A
73
C
98
A
24
C
49
74
B
99
D
25
B
50
A
–668 to
–670°F
75
C
100
B
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FE ENVIRONMENTAL SOLUTIONS
1.
Refer to the Quadric Surface (Sphere) section in the Mathematics chapter of the FE Reference
Handbook.
(x − h)2 + (y − k)2 + (z − m)2 = r2 with center at (h,k,m)
(x − 0)2 + (y − 1)2 + (z − (−2))2 = r2
x2 + (y − 1) 2 + (z + 2)2 = 81
THE CORRECT ANSWER IS: A
2.
The roots of a function are defined as points where F = 0.
In this case, divide the polynomials:
x2 + 5x + 6
x + 1 x 3 + 6 x 2 + 11x + 6
x3 + x 2
5x 2 + 11x
5x2 + 5x
6x + 6
6x + 6
0
x2 + 5x + 6 factors to (x + 2)(x + 3). Therefore, the roots of F are x = –2 and x = –3
THE CORRECT ANSWER IS: C
3.
Refer to the First-Order Linear Homogeneous Differential Equations with Constant Coefficients
section in the Mathematics chapter of the FE Reference Handbook.
General solution to y' + ay = 0 is y = Ce–at
∴ for y′ + 2y = 0  y = Ce–2t
Using the initial condition, 3 = Ce–2(0)
C=3
∴ y = 3e–2t
THE CORRECT ANSWER IS: A
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FE ENVIRONMENTAL SOLUTIONS
4.
X = ab2(c)0.5=(3.5)(2.0)2(5.2)0.5=31.9
dx/x = (0.2/3.5) + 2(0.1/2.0) + 0.5(0.3/5.2) = 0.19
dx = 0.19(x) = 0.19(31.9) = 5.9
X = 31.9 + 5.9
THE CORRECT ANSWER IS: D
5.
Refer to the Straight Line section in the Mathematics chapter of the FE Reference Handbook.
f(x) = (x – 2) –1
f’(x) = –1(x – 2) –2(1) = – ( x –2) –2
f’(–1) = –1/(–1 – 2)2 = –1/9
Slope of tangent line to f(x) at x = –1 is –1/9
(x, y) = (–1, –1/3)
y – y 1 = b(x – x 1)
y – (–1/3) = –1/9[x – (–1)]
y + 1/3 = –1/9(x) –1/9
y = –1/9(x) –4/9
THE CORRECT ANSWER IS: A
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FE ENVIRONMENTAL SOLUTIONS
6.
Refer to the Dispersion, Mean, Median, and Mode Values section in the Engineering Probability
and Statistics chapter of the FE Reference Handbook.
σ=
1
2
Σ ( x1 − μ )
Ν
2
2
2
4 (11 − 12 ) + 1(12 − 12 ) + 2 (13 − 12 ) + 1(14 − 12 )
σ=
8
2
σ = 1.118
THE CORRECT ANSWER IS: D
7.
Refer to the Laws of Probability section in the Engineering Probability and Statistics chapter of
the FE Reference Handbook.
For independent events A and B,
P(A,B) = P(A)P(B|A) = 1/4 · 1/2 = 1/8 = 0.125
THE CORRECT ANSWER IS: C
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FE ENVIRONMENTAL SOLUTIONS
8.
Refer to the Linear Regression and Goodness of Fit section in the Engineering Probability and
Statistics chapter of the FE Reference Handbook.
n=8
∑xiyi = (2)(0.3) + (3)(2.5) + (5)(2.1) + (8)(3.6) + (9)(3.1) + (11)(6.5) + (13)(5.5)
+ (15)(6.0) = 308.3
∑x2 = (2)2 + (3)2 + (5)2 + (8)2 + (9)2 + (11)2 + (13)2 + (15)2 = 698
∑xi = 2 + 3 + 5 + 8 + 9 + 11 + 13 + 15 = 66
xaverage = 66/8 = 8.25
∑yi = 0.3 + 2.5 + 2.1 + 3.6 + 3.1 + 6.5 + 5.5 + 6 = 29.6
Yaverage = 29.6/8 = 3.7
Sxy = ∑xiyi – (1/n)(∑xi)(∑yi) = 308.3 – (1/8)(66)(29.6) = 64.1
Sxx = ∑x2 – (1/n)(∑xi)2 = 698 – (1/8)(66)2 = 153.5
b = Sxy/Sxx = 64.1/153.5 = 0.42
a = yaverage – bxaverage = 3.7 – (0.42)(8.25) = 0.24
Equation of line is y = a +bx
y = 0.24 +0.42x
THE CORRECT ANSWER IS: A
9.
Refer to the Engineering Probability and Statistics chapter of the FE Reference Handbook.
THE CORRECT ANSWER IS: A
10.
Refer to the Ethics chapter of the FE Reference Handbook. The Rules of Professional Conduct
states the following:
Licensees shall undertake assignments only when qualified by education or experience in
the specific technical fields of engineering or surveying involved.
THE CORRECT ANSWER IS: C
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FE ENVIRONMENTAL SOLUTIONS
11.
Refer to the NCEES Rules of Professional Conduct, Section A.4, in the Ethics chapter of the
FE Reference Handbook.
THE CORRECT ANSWER IS: B
12.
Examinees are expected to be familiar with terms used in hazardous waste regulations. Joint and
several liability may require any party to bear the entire cost.
THE CORRECT ANSWER IS: D
13.
Examinees are expected to be familiar with the major U.S. environmental regulatory acts.
THE CORRECT ANSWER IS: D
14.
Examinees are expected to be familiar with the U.S. pollution prevention acts.
Most Preferred
Source reduction
Recycling
Treatment
Least Preferred
Disposal
THE CORRECT HIERARCHY IS SHOWN ABOVE.
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FE ENVIRONMENTAL SOLUTIONS
15.
Refer to the cash flow equations table in the Engineering Economics chapter of the FE Reference
Handbook.
$5,000
i = 10%
0
1
2
3
10
$7,500
Am = $500/yr
A

A

A = Am + P  , i, n − SV  , i, n
P

F

A

A

= 500 + 7,500  , 10%, 10 − 5, 000  , 10%, 10
P

F

= 500 + 7,500 ( 0.16275) − 5, 000 ( 0.06275)
= $1, 407 per year
THE CORRECT ANSWER IS: C
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FE ENVIRONMENTAL SOLUTIONS
16.
Refer to the cash flow equations table in the Engineering Economics chapter of the FE Reference
Handbook.
Given that demand D = 1,200 units/month
So purchase price per unit: P = 220 – 0.025 × 1,200 = 220 – 30 = $190
Thus the monthly revenue = P × D = $190 × 1,200 = $228,000 per month
Given that fixed cost = $75,000 per month
The variable cost per unit = $82, so the monthly variable costs = $82 × 1,200 = $98,400 per month
Total costs per month = fixed cost + variable costs = $75,000 + $98,400 = $173,400 per month
Profit = revenue – total costs = $228,000 – $173,400 = $54,600 per month
THE CORRECT ANSWER IS: $54,600
17.
Refer to the Breakeven Analysis section in the Engineering Economics chapter of the FE Reference
Handbook.
Tool manufacturing costs = $1.50 (5,000) = $7,500
Automated tool manufacturing costs = $0.50 (5,000) = $2,500
Annual manufacturing savings = $7,500 − $2,500 = $5,000
Additional investment for automation = $15,000 − $1,000 = $14,000
Payback = $14,000/$5,000 = 2.8 years
THE CORRECT ANSWER IS: A
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FE ENVIRONMENTAL SOLUTIONS
18.
Refer to the cash flow equations table in the Engineering Economics chapter of the FE Reference
Handbook. To compare alternatives, the same number of years must be included in the analysis.
So the leased equipment must be used for 2 years after the 4-year life cycle of Equipment B, and
the resulting cash flow will be:
The present worth of Equipment A:
PWA = –$330,000 + $71,000 × (P/A, i = 6%, N = 6) = –$330,000 + $71,000 × (4.9173)
= $19,128.30
The present worth of Equipment B:
PWB = –$215,000 + $64,000 × (P/A, I = 6%, N = 4) + $1,000 × (P/A, I = 6%, N = 2) ×
(P/F, i= 6%, N = 4)
PWB = –$215,000 + $64,000 × (3.4651) + $1,000 × (1.8334) × (0.7921)
PWB = –$215,000 + $221,766.40 + $1,452.24 = $8,218.64
Of the two, the less profitable alternative is Equipment B, and the present worth of that alternative
is $8,218.64.
THE CORRECT ANSWER IS: B
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FE ENVIRONMENTAL SOLUTIONS
19.
Refer to the cash flow equations table in the Engineering Economics chapter of the FE Reference
Handbook. The easiest way to solve this problem is to look at the present worth of each option.
The present worth values are all given by:
P = First Cost + Annual Cost × (P/A, 12%, 8) – Salvage Value × (P/F, 12%, 8)
= First Cost + Annual Cost × 4.9676 – Salvage Value × 0.4039
Then PA = $63,731
PB = $63,392
PC = $63,901
PD = $63,222
The cash flows are all costs, so the two most preferable projects, those with the lowest present
worth costs, are B and D, and the difference between them is $170.
THE CORRECT ANSWER IS: B
20.
Refer to the Population Modeling section in the Environmental Engineering chapter of the
FE Reference Handbook.
Annual rate of population growth
1 10
 61, 000 
=

 45, 000 
= 1.0309 (3.09%)
Projected population in 2010 = (61,000)(1.0309)10 = 82,689
Flow = (82,689)(160 gal/day)(10−6 Mgal/gal) = 13.2 MGD
THE CORRECT ANSWER IS: C
21.
Refer to the Population Modeling section in the Environmental Engineering chapter of the
FE Reference Handbook.
2,000 people – 1,820 people
18 people
(linear model)
k =
10
yr
 18 people 
In 30 yr, P30 = 2,000 people + 
 ( 30 yr )
yr


P30 = 2,540 people
THE CORRECT ANSWER IS: B
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FE ENVIRONMENTAL SOLUTIONS
22.
Refer to the Population Modeling section in the Environmental Engineering chapter of the
FE Reference Handbook.
Δt1 =10 years
K=
ln Pt – ln P0 ln 225,000 – ln 150,000
=
= 0.0405
10
Δt1
Δt2 = 16 years
ln Pt = ln P0 + kΔt2
= ln 225, 000 + (0.0405)(16)
= 12.97
Pt = e12.97 = 430,000
THE CORRECT ANSWER IS: C
23.
Refer to the Fate and Transport section and the Water Treatment Technologies section in the
Environmental Engineering chapter of the FE Reference Handbook.
Mass balance for first-order CSTR at steady state
dM
= QCo − QCe ± r = 0, r = −VkCe
dt
QCo − QCe − VkCe = 0
V /Q = θ =
Co − Ce
kCe
Co = 1.8 mg/L
Ce = (1 − 0.90)Co = 0.18 mg/L
k = 5 hr −1
θ=
1.8 mg/L − 0.18 mg/L
= 1.8 hr
(5 hr −1 )(0.18 mg/L)
THE CORRECT ANSWER IS: D
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FE ENVIRONMENTAL SOLUTIONS
24.
Refer to the Plug-Flow Reactor (PFR) section in the Chemical Engineering chapter of the
FE Reference Handbook.
For ideal plug flow reactor at steady state:
τ = C A0 
XA
0
C A dC A
dX A
= C A0 
0
τA
( − rA )
For first-order kinetics: −rA = KCA
C A dC A
V
1 C 
τ = = C A0 
= ln  A0 
0 KC
Q
K  C AE 
A
C 
V
Thus ln  A0  = K
Q
 C Ae 
C 
V
ln  Ae  = − K
Q
 C A0 
C Ae
V

= exp  − K 
C A0
Q


 C 
V 

η = 1 − Ae 100% = 1 − exp  − K   100%
Q 

 C A0 

K = 0.1 min−1
Hydraulic residence time = V/Q = 1 hr = 60 min
-1
[1 - e-(0.1 min )(60 min) ]
η=
= 99.8%
100
THE CORRECT ANSWER IS: C
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FE ENVIRONMENTAL SOLUTIONS
25.
Refer to the Cylindrical Pressure Vessel section in the Mechanics of Materials chapter of the
FE Reference Handbook.
The cylinder can be considered thin-walled if t < do/2. In this case, t = 12 mm and ro = do/2 = 362
mm. Thus
σt =
Pi r
t
do
r +r
350 + 362
= 356 mm
where r = i o =
2
2
σt =
(1.680 MPa)(356 mm)
= 49.8 MPa
12 mm
t
THE CORRECT ANSWER IS: B
26.
Refer to the Corrosion section in the Materials Science/Structure of Matter chapter of the
FE Reference Handbook.
Aluminum is anodic relative to copper and, therefore, will corrode to protect the copper.
THE CORRECT ANSWER IS: B
27.
Refer to the Equilibrium Constant of a Chemical Reaction section in the Chemistry chapter of the
FE Reference Handbook.
2
2
C ] [ D]
[
K=
[ A]4 [ B ]
THE CORRECT ANSWER IS: C
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FE ENVIRONMENTAL SOLUTIONS
28.
Refer to the Microbial Kinetics section in the Environmental Engineering chapter of the
FE Reference Handbook.
(
yt = L 1 − e− k1t
yt = y5 = 8 mg/L
)
L = ultimate BOD
k1 = 0.10 day−1
t = 5 days
∴ BODult =
yt
1− e
− k1 t
=
8
= 20.3 mg / L
−0.1)( 5)
(
1− e
THE CORRECT ANSWER IS: C
29.
Refer to the Atomic Bonding section in the Materials Science/Structure of Matter chapter of the
FE Reference Handbook.
These compounds are hydrocarbons, which are polymer molecules. Therefore, covalent bonding
is correct.
THE CORRECT ANSWER IS: A
30.
Refer to the Partition Coefficients section in the Environmental Engineering chapter of the
FE Reference Handbook.
Log K OWTCE = 2.3
K OWTCE = 102.3 = 200
K OWTCE = Co / C w = 200 = 15μg / L / C w
C w = 15 / 200 = 0.075 μg / L
THE CORRECT ANSWER IS: A
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FE ENVIRONMENTAL SOLUTIONS
31.
Refer to the Acids, Bases, and pH (Aqueous Solutions) section in the Chemistry chapter of the
FE Reference Handbook.
pH = 2 = – log [H+]
[H+] = 10–2 moles/L
Assuming complete dissociation of the HCl:
Moles HCl = (10–2 moles/L)(1 L) = (V)(6.0 moles/L)
V = 0.00167 L = 1.67 mL
THE CORRECT ANSWER IS: A
32.
Refer to the Radiation section in the Environmental Engineering chapter of the FE Reference
Handbook.
N = N0 exp(−0.693 t/τ)
τ=
−0.693t
ln( N /N0 )
t = 32 days
N = 150 Bq
N0 = 1,200 Bq
τ=
−0.693(32 days)
= 10.66 days
ln(150 1, 200)
THE CORRECT ANSWER IS: C
33.
Refer to the Important Families of Organic Compounds table in the Chemistry chapter of the
FE Reference Handbook. The table lists benzene as a typical arene.
THE CORRECT ANSWER IS: B
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FE ENVIRONMENTAL SOLUTIONS
34.
Refer to the Intake Rates table in the Safety chapter of the FE Reference Handbook.
For carcinogens, risk = dose × toxicity = CDI × CSF.
Given CSF = 0.80 (mg/(kg·day)–1
Ingestion in drinking water, CDI =
(CW)(IR)(EF)(ED)
(BW)(AT)
CW = concentration in water = 0.01 mg/L
IR = ingestion rate = 2.3 L/day
EF = exposure frequency = 365 days/yr
ED = exposure duration = 30 yr (given)
BW = body weight = 75 kg
AT = average time = 75 (365)
CDI =
(CW)(IR)(EF)(ED)
(BW)(AT)
(0.01 mg/L)(2.3 L/day)(365 days/yr)(30 yr)
(75kg) (75·365 days)
–4
= 1.23 × 10 mg/(kg·day)
CDI =
–1
 mg 
–4 mg
Risk = 0.80 
= 9.8 × 10–5
 × 1.23 × 10
·
·
kg
day
kg
day


THE CORRECT ANSWER IS: C
35.
Refer to the Exposure section in the Safety chapter of the FE Reference Handbook. Exposure
through food is generally limited to ingestion pathways.
THE CORRECT ANSWER IS: D
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FE ENVIRONMENTAL SOLUTIONS
36.
Refer to the Noise Pollution section in the Safety chapter of the FE Reference Handbook.
For a point source
ΔSPL = 10 log10(r1/r2)2
= 10 log10(5/500)2 = −40 dBA
SPL = 100 dBA − 40 dBA = 60 dBA
THE CORRECT ANSWER IS: C
37.
Refer to the Risk Assessment/Toxicology section in the Safety chapter of the FE Reference
Handbook.
Intake = (40 μg/m3)(0.9 m3/hr)(40 hr/wk)(10−3 mg/μg)
= 1.44 mg/wk
= 0.2057 mg/day
CDI =
0.2057 mg/day
= 0.0029 mg/(kg·day)
70 kg
THE CORRECT ANSWER IS: B
38.
Refer to the Forces on Submerged Surfaces and Center of Pressure section in the Fluid Mechanics
chapter of the FE Reference Handbook.
The mean pressure of the fluid acting on the gate is evaluated at the mean height, and the center of
pressure is 2/3 of the height from the top; thus, the total force of the fluid is:
Ff = ρg
H
3
( H ) = 1,600(9.807) (3) = 70,610 N
2
2
and its point of application is 1.00 m above the hinge. A moment balance about the hinge gives:
F (3) − F f (1) = 0
F=
Ff
3
=
70, 610
= 23,537 N
3
THE CORRECT ANSWER IS: C
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FE ENVIRONMENTAL SOLUTIONS
39.
Refer to the Energy Equation section in the Fluid Mechanics chapter of the FE Reference
Handbook.
p1
V2
p
V2
+ z1 + 1 + hm = 2 + z2 + 2 + hL
γ
γ
2g
2g
p1 p2
=
γ
γ
hL = f
V12
V22
≅0≅
2g
2g
hm = 0
( no pump )
L V2
V2
+ Σk
D 2g
2g
2
 L
V
∴ z1 − z2 =  f + Σk 
 D
 2g
Entrance k = 0.5
k = 1.0
Exit
f = 0.020
L = 60 ft
D = 6 in. = 6 in./12 in./ft = 0.5 ft
g = 32.2 ft/sec2
z1 = 103.00 ft
z2 = 101.00 ft
1/2
 z − z  
∴V =  1 2  2 g 
L
 f + Σk  
 
 D
1/2
 103.00 − 101.00 

2(32.2) 
= 

(60)
+ 1.5 
 (0.02)

0.5



V = 5.75 ft/sec
2

Q = VA = (5.75 ft sec)  π (0.5 ft)  = 1.128 cfs
4


THE CORRECT ANSWER IS: A
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FE ENVIRONMENTAL SOLUTIONS
40.
Refer to the Manning's Equation section in the Fluid Mechanics chapter of the FE Reference
Handbook.
Q=
1.486
2/3
A ( R ) S1/ 2
n
Q=
 WD 
1.486
(WD ) 

0.015
 W + 2D 
 WD 
Q = 4.430 (WD ) 

 W + 2D 
Using: W = 10 and D = 2;
Q = 112 cfs
2/3
( 0.002 )1/ 2
2/3
THE CORRECT ANSWER IS: A
41.
Refer to the Fluid Flow Machinery section in the Fluid Mechanics chapter of the FE Reference
Handbook.
200
PUMPS IN SERIES
PERFORMANCE B
PUMP CURVE
TOTAL HEAD (ft)
150
PERFORMANCE A
PUMP CURVE
100
SYSTEM CURVE
ANSWER
50
PUMPS IN PARALLEL
0
100
200
300
FLOW RATE (gpm)
THE CORRECT ANSWER IS SHOWN ON THE GRAPH.
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FE ENVIRONMENTAL SOLUTIONS
42.
Refer to the Fluid Flow Measurement section in the Fluid Mechanics chapter of the FE Reference
Handbook.
ρv2
= gh ( ρ − ρair )
2
2
∴h =
( 2 ) ≈ 0.204 m
v2
ρv 2
≈
≈
2 g ( ρ − ρair ) 2 g ( 2 )( 9.8 )
THE CORRECT ANSWER IS: D
43.
Refer to the Blowers section in the Fluid Mechanics chapter of the FE Reference Handbook.
Pw =
WRT1  P2 

Cne  P1 

0.283

− 1

( 200 hp )  sec-hp  ( 0.283)(.81)
550 ft-lb
( 2 seclb )(
 P 0.283 
=  2 
− 1
P
ft-lb

53.3
112°F + 459.69 ) °R  1 
(
lb air-°R

)

 P 0.283 
0.4137 =  2 
− 1
 P1 

P 
1.4137 =  2 
 P1 
1.4137
1
0.283
=
0.283
P2
14.7
P2 = (14.7 )( 3.399 )
P2 = 50 psia
THE CORRECT ANSWER IS: B
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FE ENVIRONMENTAL SOLUTIONS
44.
Refer to the Hydraulic Gradient (Grade Line) and Energy Line (Bernoulli Equation) sections in
the Fluid Mechanics chapter of the FE Reference Handbook.
Velocity head is not part of the hydraulic grade line but is part of the total energy line.
THE CORRECT ANSWER IS: C
45.
Refer to the Water Treatment Technologies section in the Environmental Engineering chapter of
the FE Reference Handbook.
θ = V/Q
V = ( θ)( Q ) =
(45 sec)(0.30 MGD)(106 gal/mgal)
3
(86, 400 sec /day)(7.48 gal/ft )
V each unit = V/2 = 10.44 ft3
= 20.89 ft 3
V = (πr2)Depth = π(Depth)3
1/3
V 
Depth =  
π
1/3
 10.44 ft 3 
=


π


= 1.49 ft
THE CORRECT ANSWER IS: A
46.
Refer to the Energy Equation section in the Fluid Mechanics chapter of the FE Reference
Handbook.
Static head = z2 − z1 = 199.00 − 93.00 = 106.00 ft
THE CORRECT ANSWER IS: D
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FE ENVIRONMENTAL SOLUTIONS
47.
Refer to the Head Loss Due to Flow section in the Fluid Mechanics chapter of the FE Reference
Handbook.
2
 L  v 
hL = f   

 D   2 g 
2

 200   2
hL,AB = 0.03 
 = 0.03(100)(0.062) = 0.186 ft
 
 2   2 × 32.2 
2
 200   1.3 
hL,AC = 0.03 
 = 0.03(67)(0.026) = 0.052 ft

 3   2 × 32.2 
hL,CB = hL,AB − hL,AC = 0.186 − 0.052 = 0.134 ft
Since the head loss from A to B is larger than the head loss from A to C, the head at C is greater
than the head at B. This means there is flow from C to B.
THE CORRECT ANSWER IS: B
48.
Refer to the Manning's Equation section in the Fluid Mechanics chapter of the FE Reference
Handbook.
k
Q = AR2/3 s1/2
n
k = 1 (metric units)
n = 0.014
A = πD2/4 = π(1 m)2/4 = 0.785 m2
R = D/4 = (1 m)/4 = 0.25 m
S = 0.5% = 0.005
1
∴Q = 0.014 (0.785)(0.25)2/3(0.005)1/2 = 1.57 m3/s
THE CORRECT ANSWER IS: A
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FE ENVIRONMENTAL SOLUTIONS
49.
Refer to the Euler's Equation section in the Fluid Mechanics chapter of the FE Reference
Handbook.
Flow is unsteady, and the change in pressure can be determined by Euler’s equation:
( P2 + γ • Z 2 ) − ( P1 + γ • Z1 ) = −Δx • ρ • ax
Given ΔQ/t = 0.50 m3/s per second, and
ΔQ
Q
ΔV
Q = V • A so V =
and
= t =a
A
t
A
So the acceleration in the x-direction, ax = 0.50 m3/s per second / 0.25 m2 = 2.0 m/s2.
The pipe is horizontal, so Z1 = Z2. Density ρ = 1,000 kg/m3. B is greater in the x-direction than A,
and so A is location 1, and B is location 2.
PB = PA – Δx  ρ  ax
PB = 100,000 N/m2 – (3 m)  (1,000 kg/m3)  (2.0 m/s2)
PB = 100, 000
N
N
m
2
− 6, 000
2
m
= 94, 000 N/m 2
THE CORRECT ANSWER IS: A
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FE ENVIRONMENTAL SOLUTIONS
50.
Refer to the Compressors section in the Fluid Mechanics chapter of the FE Reference Handbook.
Q = m cp (T2 – T1)
Solve for T1
–T1 = [Q/m cp] – T2
T1 = T 2 – [Q/m cp]
Substitute and solve
T 1 = 100°F – [100,000 Btu/min/(500 lb/min)(0.26 Btu/lb-°F)]
T 1 = 100°F – (769°F)
T 1 = –669°F
Since the negative is showing a loss of heat, the initial temperature = 669°F.
THE CORRECT ANSWER IS: –668°F to –670°F
51.
Refer to the Newton's Law of Cooling section in the Heat Transfer chapter of the FE Reference
Handbook.
Q = hAΔT
Q = 72(2)(150)
Q = 21, 600 W
THE CORRECT ANSWER IS: D
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FE ENVIRONMENTAL SOLUTIONS
52.
Use the Ideal Gas Law from the Thermodynamics chapter of the FE Reference Handbook:
Pv
1 1 T1 = P2 v2 T2
Given P2 P1 = 2 and T2 T1 = 1/ 2
Solving for v2
v2 =
T2 Pv
1
1 1
= v1
T1 P2
4
THE CORRECT ANSWER IS: D
53.
Refer to the Hydrology/Water Resources section in the Civil Engineering chapter of the
FE Reference Handbook.
For CN = 73 and P = 4.1 in.
S=
1,000
1,000
− 10 =
− 10 = 3.70 in.
CN
73
Q=
( P – 0.25) 2 [ 4.1 – (0.2)(3.70)]2
=
= 1.60 in.
P + 0.85
4.1+ (0.8)(3.70)
Runoff volume = Area × Q

ft 2   1.6 in. 
=  35 ac × 43.560
 


ac

  12 in. ft 
= 203,165 ft 3
THE CORRECT ANSWER IS: C
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FE ENVIRONMENTAL SOLUTIONS
54.
Refer to the Hydrology/Water Resources section in the Civil Engineering chapter of the
FE Reference Handbook.
Water demand is 35,000 people × 80
gal
Mgal 30 days 84 Mgal
= 2.8
×
=
.
capita / day
day month
month
Available water is less than demand in June, July, and August.
Month
June
July
August
Available
1.2
1.0
0.8
Demand
2.8
2.8
2.8
To Be Supplied by Reservoir
1.6 MGD × 30 days = 48 MG
1.8 MGD × 30 days = 54 MG
2.0 MGD × 30 days = 60 MG
48 MG + 54 MG + 60 MG = 162 MG
THE CORRECT ANSWER IS: C
55.
The pond is empty at time = 0.
During hour 1, the inflow rate is 16 cfs. For 3,600 sec/hr, the inflow volume is 57,600 ft3. Since
the pond was empty during the previous hour, the outflow rate is 0 cfs during hour 1, and thus the
outflow volume during hour 1 is 0. Therefore, the total volume of water in the pond at the end of
hour 1 is 57,600 ft3.
During hour 2, the inflow rate is 12 cfs (given in question statement). For the hour, the inflow
volume is 43,200 ft3. Since the pond had 57,600 ft3 at the end of the previous hour, then the outflow
rate during hour 2 is 7 cfs (from the given table). This is an outflow volume of 25,200 ft3.
Therefore, the total volume of water in the pond at the end of hour 2 is 57,600 ft3 + 43,200 ft3 –
25,200 ft3 = 75,600 ft3.
During hour 3, the inflow rate is 0 cfs. The inflow volume is 0 ft3. Since the pond had 75,600 ft3
at the end of the previous hour, then the outflow rate during hour 3 is 9 cfs (from the given table).
This is an outflow volume of 32,400 ft3 during the hour. Therefore, the total volume of water in
the pond at the end of hour 3 is 75,600 ft3 – 32,400 ft3 = 43,200 ft3.
From the given table, if the volume of water in the pond is 43,200 ft3 , then the water depth = 2 ft.
THE CORRECT ANSWER IS: A
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FE ENVIRONMENTAL SOLUTIONS
56.
Refer to the Streeter Phelps equation in the Stream Modeling section in the Environmental
Engineering chapter of the FE Reference Handbook.
tc =
k 
( k − k ) 
1
ln  2 1 − Do 2 1  
k2 − k1  k1 
k1Lo  
D=
k1Lo
 exp ( − k1t ) − exp ( − k2t )  + Do exp ( − k2t )
k2 − k1 
DO = Dsat − D
From the question statement:
k2 =
k1 =
Do =
Lo =
0.4 day−1
0.2 day−1
Dsat − DOinitial = 9.2 − 7.0 = 2.2 mg/L
15 mg/L
tc = 2.67 days
Dtc = 4.39 mg/L
DO = 9.2 − 4.39 = 4.8 mg/L
THE CORRECT ANSWER IS: C
57.
Refer to the Mass Balance section in the Environmental Engineering chapter of the FE Reference
Handbook.
Precipitation inflow = 2,000 acres × 3 ft = 6,000 acre-ft
Evapotranspiration losses = 2,000 acres × 16.2 in / 12 in/ft = 2,700 acre-ft
Seepage losses = 21 acres × 7.8 in / 12 in/ft = 13.65 acre-ft
Change in volume 1 = 6,000 – 2,700 – 13.65 = 3,286 acre-ft
3,286 acre-ft × 43,560 ft2/acre = 143 M ft3 × 7.48 gal/ft3= 1,071 Mgal
Total volume change = 1,071 – 2.5 MGD (365 days/yr) = 1,071 – 913 = 158 Mgal
Volume at end of year = 205 +158 = 363 Mgal
THE CORRECT ANSWER IS: C
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FE ENVIRONMENTAL SOLUTIONS
58.
Refer to the Rational Formula section in the Civil Engineering chapter of the FE Reference
Handbook.
Q = CiA
C = 0.40
A = 12 acres
Must use intensity, duration, frequency curve to determine i.
Duration = 1 hr = 60 min
Frequency = 10 yr
∴ i = 1.8 in./hr
Q = (0.40)(1.8)(12) = 8.64 cfs
THE CORRECT ANSWER IS: A
59.
Refer to the Hydrology/Water Resources section in the Civil Engineering chapter of the
FE Reference Handbook.


gal
6
Qmax = (1.5) 100
 (300, 000 capita) = 45 ×10 gal/day
day − cap 

Vmax = Qmax × t = (45 × 106 gal/day)(1 day) = 45 × 106 gal
Vfire = (3, 000 gal / min)(8 hr)(60 min/hr) = 1.44 × 106 gal
Vdaily = Vmax + Vfire
= 45 × 106 gal + 1.44 × 106 gal
= 46.4 × 106 gal
THE CORRECT ANSWER IS: C
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FE ENVIRONMENTAL SOLUTIONS
60.
Examinees are expected to be familiar with term eutrophication. Eutrophication is stimulated by
plant nutrients N and P. Limits on total N and total P are usually specified to address eutrophication
concerns.
THE CORRECT ANSWER IS: C
61.
Refer to the Rational Formula section in the Civil Engineering chapter of the FE Reference
Handbook.
Q = CIA
A =
10
Q
=
= 2.26 acres
CI 0.65 × 6.8
Catchment Area = 2.26 acres × 43,560 ft 2 / acre = 98,400 ft 2
A = 100 × L
L =
98, 400
= 984 ft
100
THE CORRECT ANSWER IS: A
62.
Examinees are expected to be familiar with the properties of facultative bacteria. Facultative
bacteria can use oxygen as an electron acceptor if available. If oxygen is not available, facultative
bacteria can use other electron acceptors such as nitrate or organic compounds.
THE CORRECT ANSWER IS: D
63.
Refer to the overflow rate in the Clarifier section in the Environmental Engineering chapter of the
FE Reference Handbook.
A = L × W (surface area)
Overflow rate = Q/A
A = (3 cells)(10 ft)(10 ft) = 300 ft2
Overflow rate =
( 2 MGD) (106 gal/Mgal )
(300 ft ) (1, 440 min/day)
2
= 4.63 gpm/ft 2
THE CORRECT ANSWER IS: B
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FE ENVIRONMENTAL SOLUTIONS
64.
Refer to the equation for hydraulic residence time in the Clarifier section in the Environmental
Engineering chapter of the FE Reference Handbook.
Residence time =
volume
flow rate
 ( 65 ft )2

Clarifier volume = 2 ×  π
(12 ft ) = 79,639 ft 3
4




 5 MGD   106 gal / Mgal 
Flow = 
= 27,852 ft 3 / hr
 7.48 gal / ft 3  24 hr / day 



Residence time =
79,639 ft 3
27,852 ft 3 / hr
= 2.86 hr
THE CORRECT ANSWER IS: C
65.
Examinees are expected to be familiar with trihalomethanes as a by-product of disinfection.
Trihalomethanes are by-products of disinfection. Chlorine reacts with organic precursors to form
trihalomethanes.
THE CORRECT ANSWER IS: C
66.
Refer to the Water Treatment Technologies section in the Environmental Engineering chapter of
the FE Reference Handbook. A steady-state mass balance for suspended solids around an activated
sludge final clarifier yields the following equation:
(Qo + QR )X A = Qe X e + QR X w
where Xw is the return sludge suspended solids concentration. Thus:
Xw =
(Qo + QR )X A − Qe X e
QR
mg 

 mg 
(12 MGD + 10 MGD)  3,000
 − (12 MGD)  0

L 

 L 
Xw =
10 MGD
Xw = 6,600 mg/L
THE CORRECT ANSWER IS: B
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FE ENVIRONMENTAL SOLUTIONS
67.
Refer to the Wastewater Treatment and Technologies section in the Environmental Engineering
chapter of the FE Reference Handbook.
Primary sludge dry weight
Weight = QSoη = (1.2 MGD)(250 mg/L)(8.34 (lb/mgal)/mg/L)(0.60) = 1,501 lb/day
Wet weight =
Volume =
1,501 lb dry/dry
= 37,530 lb/day
0.04 lb dry/lb wet
37,530 lb/day
= 601 ft 3 /day
3
62.4 lb/ft
THE CORRECT ANSWER IS: A
68.
Examinees are expected to be familiar with various treatment technologies.
Treatment Objective
A. Remove suspended solids
Treatment Technology
2. Media filtration
B. Reduce dissolved chemicals
3. Biologically active filtration
C. Degrade trace organics
1. UV/H2O2
THE CORRECT ANSWERS ARE SHOWN.
69.
1. Calculate the metric tons of CO2 emitted.
CO2 = (12,395 kW-h/year) × (900 lb CO2/MW-h) × (1 MW-h/1,000 kW-h) × (1 metric ton/2,204.6
lb) = 5.06 metric tons CO2/year
2. Account for the losses due to transmission and distribution.
CO2 = (5.06 metric tons CO2/year) × [1/(1 – 0.08)] = 5.5 metric tons CO2/year
Answer = 5.5 metric tons CO2/year
THE CORRECT ANSWER IS: B
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FE ENVIRONMENTAL SOLUTIONS
70.
Refer to the Fate and Transport section in the Environmental Engineering chapter of the
FE Reference Handbook.
ΣQiCi = QT Cave
Inflow = Q4 = 0.1(4.0 + 0.8 + 0.2) = 0.5 million liters per day
C4 = 0
Cave =
 Qi Ci 4(200) + 0.8(300) + 0.2(500)
=
= 207 mg/L
(4.0 + 0.8 + 0.2 + 0.5)
QT
THE CORRECT ANSWER IS: B
71.
Examinees are expected to be familiar with surface water treatment facility technologies. The
correct order is shown below.
1.
2.
3.
4.
5.
Rapid mix
Flocculation
Sedimentation
Filtration
Disinfection
THE CORRECT ANSWER IS SHOWN.
72.
Refer to the Lime Soda Softening Equations section in the Environmental Engineering chapter of
the FE Reference Handbook.
Hardness is associated with Ca2+ and Mg2+, not Na+. Convert concentrations to mg/L as CaCO3:
 70 mg L   2meq   mg CaCO3 
Ca2+: 

 = 175 mg L
  50
meq

 40 mg mmole   mmole  


 mg CaCO3 
30 mg L
Mg2+: 
 ( 2 meq mmole )  50
 = 123.456 mg L
meq


 24.3 mg mmole 
Total Hardness = 175 + 123 = 298mg L as CaCO3
THE CORRECT ANSWER IS: 297–300 mg/L as CaCO3
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FE ENVIRONMENTAL SOLUTIONS
73.
Refer to the Wastewater Treatment Technologies section in the Environmental Engineering
chapter of the FE Reference Handbook.
Organic loading rate (F:M) =
lb BOD 5 /day to aeration
Qo S o
=
lb SS in aeration
Vol X A
Qo = 1.2 MGD
So = 165 mg/L
Vol = (90 ft)(30 ft)(12 ft)(7.48 gal/ft3)(10−6 Mgal/gal) = 0.242 Mgal
XA = 4,000 mg/L
Organic loading =
(1.2 )(165) = 0.204 lb BOD5 /day
lb SS
( 0.242 )( 4,000 )
THE CORRECT ANSWER IS: C
74.
Refer to the Incomplete Combustion section in the Thermodynamics chapter of the FE Reference
Handbook.
Carbon monoxide poses a risk of an elevated emission rate for combustion sources with
operational anomalies resulting in concentrations above safe levels in living spaces.
Radon is an indoor air pollutant that is a naturally emitted decay product from geological
formations.
Carbon dioxide is a by-product of combustion under normal operating conditions and is less of a
concern to human health for indoor air quality.
Elevated formaldehyde emissions are not a risk from operation; formaldehyde emissions are
typical of composite wood and laminate and not from kerosene heaters.
THE CORRECT ANSWER IS: B
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FE ENVIRONMENTAL SOLUTIONS
75.
Refer to the First Law of Thermodynamics section in the Thermodynamics chapter of the FE
Reference Handbook.
See the equation that refers to conservation of mass, i.e. the sum of the mass entering and exiting
a system being equal. Note that conservation of mass only applies in an open system if there are
no losses in the system. If there are no losses, then  min =  mexit .
This equation extends to the mass rate of flow (i.e. mass/time).
Mass is conserved.
Input mass = output mass = 75 lb/min
THE CORRECT ANSWER IS: C
76.
Refer to the Incomplete Combustion section in the Thermodynamics chapter of the FE Reference
Handbook.
A slight increase in air-to-fuel ratio provides more complete combustion, thus:
CO
→
VOC →
N
→
CO2
CO2
NOx
THE CORRECT ANSWER IS: D
77.
Refer to the Psychrometrics section in the Thermodynamics chapter of the FE Reference
Handbook.
The definition of relative humidity:
φ=
pv
vapor pressure
=
pg saturation pressure
THE CORRECT ANSWER IS: A
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FE ENVIRONMENTAL SOLUTIONS
78.
Examinees are expected to be familiar with emission control technologies. SOx is removed by
liquid scrubbing at a high pH. Lime may be added to raise the pH.
THE CORRECT ANSWER IS: A
79.
Baghouses are appropriate technology for the removal of air pollutants that are particulates. The
correct answers are particulate pollutants listed in the Baghouse section in the Environmental
Engineering chapter of the FE Reference Handbook. The incorrect options (carbon monoxide,
formaldehyde, and radon) are pollutants that are in the gas phase and would not be removed by a
baghouse.
THE CORRECT ANSWERS ARE: A, C, E, AND F
80.
Refer to the Mass Balance section in the Environmental Engineering chapter of the FE Reference
Handbook.
Mass balance at steady state:
Input − Output + Generation = Accumulation
QCo − QC + E = 0
C=
QCo + E
Q
 Air Changes 
3
Q =  80
 (50 m × 20 m × 10 m) = 800, 000 m /hr
hr


Co = 0
E = (1 kg/hr ) (109 μg/kg) = 109 μg/hr
C=
109 μg/hr
3
800, 000 m /hr
= 1, 250 μg/m3
THE CORRECT ANSWER IS: C
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FE ENVIRONMENTAL SOLUTIONS
81.
Refer to the Air Pollution section in the Environmental Engineering chapter of the FE Reference
Handbook.
C (μg/m3 ) =
 1 y 2  
 1 ( z − H )2 
 1 ( z + H ) 2  
Q
exp  −
exp
exp
−
+

−

 2 σ 2 
 2 σ 2  
 2 σ 2y  
2πU σ y σ z
z
z






Q = 100 μg/s
U = 3 m/s
y=0m
H=0m
z=0m
For neutral stability (class D) and x = 0.1 km,
σy = 8 m
σz = 4.5 m
C=
100
exp (0) {exp (0) + exp(0)} = 0.295 μg/m3
(2π)(3)(8)(4.5)
THE CORRECT ANSWER IS: A
82.
The following table shows how much of each component is recovered by recycling and how much
remains as solid waste. The basis is 100 lb of original MSW.
Component
Paper
Yard
Food
Plastic
Metal
Wood
Glass
Other
Mass
(lb)
35
20
10
9
8
5
5
8
100
Recovered
(lb)
14
16
0
0
1.6
0
1.5
0
33.1
Remaining
(lb)
21
4
10
9
6.4
5
3.5
8
66.9
100 − 66.9
×100% = 33.1%
100
Reduction
=
THE CORRECT ANSWER IS: B
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FE ENVIRONMENTAL SOLUTIONS
83.
P + Qin – Qout + QGrnd – ES – TS – I = ΔS
Transpiration = 0 due to lack of plants on pond margins
Infiltration = 0 due to soils
QGrnd = 0 due to soils
Flowin = (0.450 ft3/sec)*(3,600 sec/hr)*(24 hr/day)*(30 day/mo) = 1.1664 × 106 ft3/mo
(1.1664 × 106 ft3/mo)*(1 acre/43,560 ft2 *(1/10 acre)*(1 mo) = 2.678 ft
Rain = 2.50 in./(12 in./ft) = 0.2083 ft
Flowout = (0.10 ft3/sec)*(3,600 sec/hr)*(24 hr/day)*(30 day/mo) = 259,200 ft3/mo
(259,200 ft3/mo)*(1 acre/43,560 ft2*(1/10 acre)*(1 mo) = 0.595 ft
2.678 ft + 0.2083 ft – 0.5950 ft – ES = ΔS = 514.8 – 512.9 = 1.9 ft
ES = 2.678 ft + 0.2083 ft – 0.595 ft – 1.9 = 0.3909 ft
E = 0.391 ft ≈ 4.7 in.
THE CORRECT ANSWER IS: 4.2–5.2 in.
84.
Determine mass of cover.
 1 lb cover 
Cover = (50, 000 lb refuse) 
 = 16,667 lb
 3 lb refuse 
Total mass fill (10-yr life)
Fill = (50,000 lb/day refuse + 16,667 lb/day cover)(365 days/yr)(10 yr) = 2.433 × 108 lb
2.433 × 108 lb
= 243, 000 yd 3
Volume =
3
1, 000 lb/yd
THE CORRECT ANSWER IS: B
85.
Refer to the Hazardous Waste Compatibility Chart in the Safety chapter of the FE Reference
Handbook.
THE CORRECT ANSWER IS: D
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FE ENVIRONMENTAL SOLUTIONS
86.
Refer to the Sampling and Monitoring section in the Environmental Engineering chapter of the FE
Reference Handbook.
Confidence level (CL) = 80%
Power = 90%
Standard deviation of sample (s) = 0.1
Sample average ( ̅ ) = 0.4
Minimum detectable relative difference (MDRD) = 20%
1. Calculate the coefficient of variation (CV)
CV = (100 * s) / ̅
CV = (100 * 0.1)/0.4
CV = 25%
2. Use the data quality objectives (DQO) for sampling soils (FE Reference Handbook) tables to
determine the number of samples required.
The intersection of CV = 25%, Power = 90%, Confidence Level = 80%, MDRD = 20% results in
eight samples.
THE CORRECT ANSWER IS: B
87.
Refer to the Hazard Assessment section in the Safety chapter of the FE Reference Handbook.
According to the fire/hazard diamond, the diamond in the northern, eastern, southern, and western
positions represent the flammability, reactivity, special hazards, and health hazards, respectively.
The northern position represents the key position for this case, and a value of 2 in this position
represents a material that will ignite if moderately heated.
2
2
2
2
THE CORRECT ANSWER IS SHADED ABOVE.
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FE ENVIRONMENTAL SOLUTIONS
88.
Collection truck volume = vehicle 8 yd 3 ×
0.25 ton
yd 3
= 2 tons/collection truck
16 tons
= 8 collection trucks
2 tons/collection truck
THE CORRECT ANSWER IS: 8
89.
Refer to the Geotechnical section in the Civil Engineering chapter of the FE Reference Handbook.
Porosity = Vvoid /Vtotal = 0.40
Assume Vtotal = 1 ft3
Vvoid = (0.40)(1 ft3) = 0.40 ft3
Vsolid = 1 − Vvoid = 0.60 ft3
Wt = (0.40 ft 3 ) γ water + (0.60 ft 3 )(SG sand )( γ water )
= ( 0.40 ) + ( 0.60 )( 2.65 )  62.4 lb/ft 3 = 124 lb/ft 3
THE CORRECT ANSWER IS: B
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FE ENVIRONMENTAL SOLUTIONS
90.
Refer to the Geotechnical section in the Civil Engineering chapter of the FE Reference Handbook.
Q = 25 cm3/day
dh = 4
i = dL
A=
π 2 π
D = (4) 2
4
4
te = 1
K=
Q
25 cm3 /day
=
i A tE
π 
4 42 1
4 
K = 0.497
THE CORRECT ANSWER IS: B
91.
Refer to Dupuit's formula in the Civil Engineering chapter of the FE Reference Handbook.
THE CORRECT ANSWER IS: A
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FE ENVIRONMENTAL SOLUTIONS
92.
Refer to the Air Stripping section in the Environmental Engineering chapter of the FE Reference
Handbook.
Z = HTU × NTU
Calculate NTU:
 100 m 3·s air 
RS = ( 0.5 ) 
 = 50
3
·
1
m
s
water


NTU =
 ( C /C )( RS – 1) +1  50  ( 275/2 )( 49 ) +1 
RS
ln  in out
 = ln 

RS – 1 
RS
50
 49 

NTU = 5.00
Calculate HTU:
HTU =
HTU =
L
M W KLa
10 kmol / s·m 2
= 12.93 m
kmol   50   1 hr 

 55.6 3   h   3,600 s 

m  

Z = HTU × NTU
5 × 12.93 m = 64.7 m
THE CORRECT ANSWER IS: B
93.
Refer to the Well Drawdown section in the Civil Engineering chapter of the FE Reference
Handbook.
Recharge associated with an unconfined aquifer would generally have a shorter response time.
THE CORRECT ANSWER IS: A
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FE ENVIRONMENTAL SOLUTIONS
94.
Refer to Darcy's law in the Civil Engineering chapter of the FE Reference Handbook.
Darcy's Equation
Q = – (KA)(dh/dx)
V = Q/A = –K Δh/Δx
K = 5 ft/day
Δh = 100.85 ft – 101.00 ft = –0.15 ft
Δx = 250 ft
−(0.15 ft)
V = −(5 ft/day)
= 0.003 ft/day
(250 ft)
THE CORRECT ANSWER IS: B
95.
Refer to Dupuit's formula in the Civil Engineering chapter of the FE Reference Handbook.
Q=
(
πk h22 − h12
r1 =
=
r 
ln  2 
 r1 
)
r2
 πk (h22 − h12 ) 
exp 

Q


30ft
 π(5 ×10 ft/sec)(202 −102 )ft 2 
exp 

1×10−2 ft 3/sec


–5
r1 = 0.27 ft
d1 = 0.54 ft = 6.5 in.
THE CORRECT ANSWER IS: D
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FE ENVIRONMENTAL SOLUTIONS
96.
Refer to Darcy's law in the Civil Engineering chapter of the FE Reference Handbook.
For T = K · b K =
K=
T
, given T = 1.6 × 10–2 m2/s and b = 45 m
b
1.6 × 10−2 m2 /s
= 3.56 × 10−4 m / s
45 m
Darcy’s Law
dh
for dh = – 0.37 m and dx = 50 m
dx
 −0.37 m 
3
3
Q = – (3.56 × 10–4 m/s)(675 m2) 
 = 0.001778 m /s × 86,400 sec/day = 153.6 m /day
50
m


Q = –KA
= 150 m3/day
THE CORRECT ANSWER IS: B
97.
Refer to the Municipal Solid Wastes section and the Fate and Transport section in the
Environmental Engineering chapter of the FE Reference Handbook.
The overall energy value is the sum of the energy values from each component weighted on a mass
basis.
Component
Food waste
Paper
Plastic
Wood
Total
Weight
(lb)
20
40
10
30
100
Component Energy Value
(Btu/lb)
2,500
6,500
12,000
7,500
Total Energy Value
(Btu)
50,000
260,000
120,000
225,000
655,000
The energy value of the entire sample (Btu/lb) is 655,000 Btu/100 lb = 6,550 Btu/lb.
THE CORRECT ANSWER IS: 6,400–6,600 Btu/lb
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FE ENVIRONMENTAL SOLUTIONS
98.
There are no gas phase emissions associated with the operation of a wind turbine. All other items
would typically be evaluated during the study for the localized impacts of the wind turbine
operation.
THE CORRECT ANSWER IS: A
99.
Determine energy input rate
Btu/hr  6

(1,000 MW)  3.413
 (10 W/MW)
Power
W 

Input =
=
= 1.034 ×1010 Btu/hr
η
0.33
(
)(
)
NO x = 0.6 lb/106 Btu 1.034 ×1010 Btu/hr ( 24 hr/day ) = 1.49 ×105 lb/day
THE CORRECT ANSWER IS: D
100.
Write a balanced chemical reaction for the combustion (oxidation) of benzene
C6H6 + 7.5 O2 = 6 CO2 + 3 H2O
1. Determine the mass ratio of the reactants (note the balanced reaction provided the moles of
each reactant).
Reactant
Benzene (C6H6)
Oxygen (O2)
Mole ratio
Molecular Weight
(g/mole)
Mass
Ratio
1
7.5
78
32
78
240
2. Calculate the kg of oxygen required.
Mass of oxygen required = 3 kg C6H6 * 240 kg O2/78 kg C6H6 = 9.2 kg O2
THE CORRECT ANSWER IS: B
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