60 problems from all 15 areas of the specification, in proportion to the published ranges. Most problems list what is given and what to find, and every solution opens step by step once you answer.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Environmental
practice problems60
areas covered15
questions on the real exam110
Checked
How the set is built
Problems per area follow the midpoint of each published range, so the two largest areas, fluid mechanics and hydraulics and water and wastewater, get the most problems and the smallest areas get two or three. The shared-area problems (mathematics, statistics, ethics, economics) are a separate set from the ones on the FE practice test page and engineering economics, so working both pages gives you new problems. The area chips over problem 1 filter the set.
60 problems, every option explained
0 / 60 answered · 0 correct
Question 1 of 60 · Fundamental Principles · 7–11 items
Two aerated lagoons operate in series as completely mixed basins. The first holds 20,000 m3 and the second 10,000 m3; the flow is 2,000 m3/day, the influent BOD is 200 mg/L, and BOD removal is first order with k = 0.25 day−1 in both. What is the effluent BOD from the second lagoon?
Given
Lagoon 1 volume = 20,000 m3
Lagoon 2 volume = 10,000 m3
Flow Q = 2,000 m3/day
Influent BOD = 200 mg/L
Rate constant k = 0.25 day−1
Find
Effluent BOD from lagoon 2 (mg/L)
Solution
Each completely mixed basin divides the concentration by (1 + kτ) (Fundamental principles: reactors in series).
4.70 mg/L treats the 15 days of total detention as plug flow, e−0.25 × 15; the lagoons are completely mixed.
24.2 mg/L splits the 15 days evenly (7.5 days each); the lagoons have different volumes, 10 and 5 days.
τ1 = 20,000/2,000 = 10 d and τ2 = 5 d; C = 200/[(1 + 2.5)(1 + 1.25)] = 200/7.875 = 25.4 mg/L.
42.1 mg/L lumps both lagoons into one 15-day CSTR; splitting the volume into two mixed basins removes more.
57.1 mg/L stops after the first lagoon (200/3.5).
Question 2 of 60 · Fluid Mechanics and Hydraulics · 12–18 items
A cylindrical buoy with a diameter of 2 m and height of 1.5 m is floating in seawater (density = 1025 kg/m³). What is the mass of the buoy if it is floating with 0.5 m above the water surface?
Given
Buoy diameter D = 2 m
Buoy height H = 1.5 m
Seawater density ρ = 1025 kg/m3
Freeboard above water = 0.5 m
Find
Mass of the buoy m (kg)
Solution
A floating body displaces its own mass of fluid (Fluid mechanics: buoyancy).
Vdisp = π(1.0 m)2 × (1.5 − 0.5) m = 3.14 m3
m = ρV = 1,025 kg/m3 × 3.14 m3 = 3,220 kg
12,900 kg uses the 2 m diameter as the radius, which quadruples the area.
6,440 kg adds the 0.5 m freeboard to the 1.5 m height (2.0 m draft) instead of subtracting it.
4,830 kg uses the full 1.5 m height as if the buoy were completely submerged.
Draft = 1.0 m, displaced volume = π × 1.02 × 1.0 = 3.14 m3, and mass = 1,025 × 3.14 = 3,220 kg.
Question 3 of 60 · Mathematics · 5–8 items
A survey point in a local coordinate system has coordinates x = 40.00 m, y = 10.00 m. The local system is rotated 30° counterclockwise relative to the global North, and the local origin is located at global coordinates (100.00, 200.00) m. What are the coordinates of the point in the global system?
Given
Local coordinates x = 40.00 m, y = 10.00 m
Rotation = 30° counterclockwise
Local origin in global system = (100.00, 200.00) m
Find
Global coordinates of the point (m)
Solution
Rotate the local coordinates by +30°, then translate by the local origin (Mathematics: coordinate transformation).
X = X0 + x cos θ − y sin θ = 100 + 34.64 − 5.00 = 129.6 m
Y = Y0 + x sin θ + y cos θ = 200 + 20.00 + 8.66 = 228.7 m
(29.6, 28.7) is the rotated point before the translation; the local origin (100.00, 200.00) still has to be added.
(140.0, 210.0) adds the translation but skips the rotation.
(139.6, 188.7) rotates clockwise (the sine terms with the opposite sign), the inverse of the stated rotation.
X = 100 + 40 cos 30° − 10 sin 30° = 129.6; Y = 200 + 40 sin 30° + 10 cos 30° = 228.7.
(111.3, 239.6) rotates by 60°, reading the 30° from the wrong reference axis.
Question 4 of 60 · Probability and Statistics · 4–6 items
An analyst is fitting a piecewise linear regression model that assumes the slope changes at a known breakpoint xb, while the left segment keeps the baseline slope β1 and the function stays continuous at xb. Which term must be added to the baseline model y = β0 + β1x? (I(·) is the indicator function.)
Solution
A linear spline (hinge) term (x − xb)+ changes the slope at xb while keeping continuity (Probability and statistics: regression).
A squared term bends the whole curve into a parabola; the model is no longer piecewise linear.
The hinge term is zero up to xb and grows as β2(x − xb) beyond it, so the slope becomes β1 + β2 with no jump at xb.
An indicator alone adds a step of β2 at xb: a jump in level, not a change in slope.
Without the indicator the term acts on both sides, changing the slope everywhere instead of only past xb.
This adds a global slope change plus a step, which breaks continuity at xb.
Question 5 of 60 · Ethics and Professional Practice · 5–8 items
Time
Release Amount (lb)
08:00 (Day 1)
30
14:00 (Day 1)
40
20:00 (Day 1)
20
02:00 (Day 2)
20
A facility releases a substance with a Reportable Quantity (RQ) of 100 lb. Releases occur as shown in the log. At what time is the immediate notification requirement triggered?
Given
Reportable Quantity RQ = 100 lb
08:00 (Day 1) release = 30 lb
14:00 (Day 1) release = 40 lb
20:00 (Day 1) release = 20 lb
02:00 (Day 2) release = 20 lb
Find
Time the notification requirement is triggered
Solution
A release of a reportable quantity or more within any 24-hour period requires immediate notification.
30 + 40 + 20 + 20 = 110 lb ≥ 100 lb within 18 h → notify at 02:00 Day 2
The 30 lb release at 08:00 is well below the 100 lb reportable quantity.
By 14:00 the cumulative release is 70 lb, still under 100 lb.
By 20:00 the total is 90 lb; the reportable quantity has not yet been reached.
The 02:00 release brings the total within 24 hours (since 08:00 Day 1) to 110 lb, at or above the 100 lb RQ, so notification is due then.
Releases are summed over any 24-hour period; 110 lb within 18 hours exceeds the RQ.
Question 6 of 60 · Ethics and Professional Practice · 5–8 items
A facility receives a Notice of Violation (NOV) requiring a response within 30 days. Which response strategy best satisfies regulatory requirements?
Solution
A notice of violation is answered with substance: cause, corrective action and a compliance schedule.
A bare promise lacks the root cause, corrective actions and dates that regulators expect.
An NOV response is expected to explain the cause, the corrective actions taken or planned, and a schedule to return to compliance.
Contesting alone, without addressing the conditions found, leaves the requested response unmet.
Waiting for another inspection lets the 30-day deadline pass with no response and the violation uncorrected.
Requesting a hearing does not supply the corrective-action plan the notice asks for, so the response would still be incomplete.
Question 7 of 60 · Engineering Economics · 5–8 items
A hazardous waste landfill requires a protective cap replacement every 15 years indefinitely. The first replacement will occur 15 years from today, and each replacement costs $250,000 at the time it occurs. At an interest rate of 5%, calculate the capitalized cost (present worth) of this replacement schedule.
Given
Replacement interval = 15 years, indefinitely
First replacement at year 15
Cost per replacement = $250,000
Interest rate i = 5%
Find
Capitalized cost
Solution
Capitalized cost of a perpetual periodic cost: convert to an annual amount with (A/F), then divide by i (Engineering economics).
A = 250,000 × 0.05/[(1.05)15 − 1] = $11,586
P = A/i = 11,586/0.05 = $231,711
$5,000,000 divides the $250,000 by i as if it were spent every year.
$120,254 is the present worth of the first replacement only; the replacements continue forever.
$11,586 is the equivalent annual amount, F(A/F, 5%, 15); it still has to be capitalized by dividing by i.
$481,711 uses (A/P) instead of (A/F), which adds an immediate replacement at time zero.
Question 8 of 60 · Engineering Economics · 5–8 items
A small business repays a $20,000 environmental compliance loan over 5 years with equal monthly payments at a nominal annual interest rate of 12%, compounded monthly. What is the principal portion of the 20th payment?
Given
Loan = $20,000
Term = 5 years, equal monthly payments
Nominal rate = 12% per year compounded monthly
Payment number = 20
Find
Principal portion of payment 20
Solution
Level-payment amortization: the principal share grows geometrically at the periodic rate (Engineering economics).
A = 20,000 × 0.01(1.01)60/[(1.01)60 − 1] = $444.89
principal20 = (444.89 − 200)(1.01)19 = $296
$445 is the whole monthly payment.
A = 20,000 × (A/P, 1%, 60) = $444.89; principal in payment k = (A − iP)(1 + i)k−1 = 244.89 × 1.0119 = $296.
$149 is the interest portion of payment 20, not the principal.
$333 is 20,000/60, a straight-line principal split; level-payment loans repay principal slowly at first.
$245 is the principal portion of the first payment; it grows by 1% each month.
Question 9 of 60 · Ethics and Professional Practice · 5–8 items
An engineer managing a site remediation project discovers a ruptured leachate collection pipe that is releasing untreated fluid into a local creek. The engineer's team stops the leak within minutes. According to standard professional and regulatory obligations, what is the engineer's immediate priority?
Solution
Release to the environment: report promptly to the authorities; analysis and paperwork follow (Ethics: public safety obligations).
Untreated leachate is known to be contaminated; waiting days for lab results delays the response.
A release of untreated leachate to surface water must be reported promptly; stopping the leak does not remove the duty to report, and protecting the public comes first.
An estimate is enough for the initial report; delaying notification to refine the volume defeats its purpose.
The duty to protect the public outranks the client's interest in controlling the message.
Root cause analysis follows later; it does not come before notification.
Question 10 of 60 · Fundamental Principles · 7–11 items
A water treatment plant has a maximum design capacity of 20 MGD and currently processes a maximum day demand of 12 MGD. The demand is projected to grow by 4.0% per year, compounded annually. After how many full years of growth will the demand first exceed the plant's design capacity?
Given
Plant capacity = 20 MGD
Current maximum day demand = 12 MGD
Growth rate = 4.0 %/yr
Find
Full years until demand exceeds capacity
Solution
Solve the compound-growth equation, then take the first full year past capacity.
12 × (1.04)t = 20 → t = ln(1.667)/ln(1.04) = 13.02 years
year 13: 19.98 MGD; year 14: 20.78 MGD > 20 MGD
12 years treats 4.0% as a continuous rate and truncates t = ln(1.667)/0.04 = 12.8; the growth here is compounded annually.
13 years rounds t = 13.02 down; after 13 years demand is 19.98 MGD, still below 20 MGD.
(1.04)t = 20/12 gives t = 13.02, so demand first exceeds 20 MGD after 14 full years (20.78 MGD).
16 years assumes linear growth of 0.48 MGD per year and truncates 16.7; the growth is compound.
17 years is the first full year above capacity under linear growth, 0.48 MGD per year, not compound growth.
Question 11 of 60 · Environmental Chemistry · 7–11 items
A reaction rate constant k is determined to be 1.5 d−1 at 25°C. If the activation energy Ea is 50 kJ/mol, what is the estimated rate constant at 35°C? (Universal Gas Constant R = 8.314 J mol−1 K−1)
Given
Rate constant k1 = 1.5 d−1 at 25 °C
New temperature = 35 °C
Activation energy Ea = 50 kJ/mol
R = 8.314 J/(mol·K)
Find
Rate constant at 35 °C (d−1)
Solution
Two-temperature Arrhenius equation with absolute temperatures (Chemistry: reaction rate).
6.8 d−1 uses 100.655 instead of e0.655, treating the natural-log Arrhenius form as base 10.
23.2 d−1 uses R = 1.987 cal/(mol·K) with Ea in joules, which makes the exponent 4.2 times too large.
Question 12 of 60 · Health Hazards and Risk Assessment · 4–6 items
A risk assessment determines a No Observed Adverse Effect Level (NOAEL) of 100 mg/kg-day for a chemical. The Estimated Exposure Dose (EED) for the population is 0.5 mg/kg-day. The derivation of the Reference Dose requires a standard uncertainty factor of 100 (10 for interspecies and 10 for intraspecies). Calculate the Margin of Exposure (MOE) and determine if the risk is considered acceptable.
Given
NOAEL = 100 mg/(kg·d)
Estimated exposure dose = 0.5 mg/(kg·d)
Uncertainty factor = 100
Find
MOE and acceptability
Solution
Margin of exposure = NOAEL ÷ exposure, compared with the combined uncertainty factor (Health hazards: risk characterization).
MOE = 100/0.5 = 200 > 100 → acceptable
0.005 inverts the ratio (exposure/NOAEL); the MOE puts the NOAEL in the numerator.
MOE = NOAEL/exposure = 100/0.5 = 200, which exceeds the required margin of 100 (the combined uncertainty factor), so the risk is acceptable.
200 is the right MOE, but a margin above the 100 benchmark means adequate protection, not unacceptable risk.
1.0 mg/kg-day is NOAEL/UF, the reference dose, not the margin of exposure.
0.005 is the inverted ratio, and comparing it with 100 then gives the wrong verdict as well.
Question 13 of 60 · Health Hazards and Risk Assessment · 4–6 items
The inhalation unit risk for a contaminant is 4.0 × 10−6 (µg/m3)−1. Assuming a standard adult body weight of 70 kg and an inhalation rate of 20 m3/day, calculate the corresponding slope factor in (mg/kg-day)−1.
Given
Inhalation unit risk = 4.0 × 10−6 (μg/m3)−1
Body weight BW = 70 kg
Inhalation rate IR = 20 m3/day
Find
Slope factor ((mg/kg-day)−1)
Solution
Unit risk per µg/m3 → slope factor per mg/kg-day (Health hazards: dose-response).
1.4 × 101 converts µg to mg with 106 instead of 103.
Question 14 of 60 · Environmental Chemistry · 7–11 items
A closed system at 25°C (298 K) contains 100 m³ of air and 10 m³ of water. A volatile organic compound with a Henry's law constant of 2.0 × 10−3 atm m3 mol−1 is introduced into the system. Assuming equilibrium is reached, calculate the mass fraction of the contaminant in the air phase. Use R = 8.206 × 10−5 atm · m3 mol−1 K−1.
Given
Temperature = 25 °C (298 K)
Air volume = 100 m3
Water volume = 10 m3
Henry's constant H = 2.0 × 10−3 atm·m3/mol
R = 8.206 × 10−5 atm·m3/(mol·K)
Find
Mass fraction in air
Solution
Mass in each phase = concentration × volume; at equilibrium Cair = H′·Cwater with H′ = H/RT.
H′ = 2.0 × 10−3 ÷ (8.206 × 10−5 × 298) = 0.0818
fair = H′Va/(H′Va + Vw) = 8.18/(8.18 + 10) = 0.45
H′ = H/RT = 0.0818; air holds 0.0818 × 100 = 8.18 units per 10 in water, so the air fraction is 8.18/18.18 = 0.45.
0.55 is the fraction left in the water phase, not in air.
0.82 divides the air term (8.18) by the water term (10) alone instead of by the total 18.18.
0.91 uses T = 25 instead of 298 K in RT, which makes the dimensionless constant about 12 times too large.
0.99 inverts the dimensionless constant (RT/H = 12.2), which puts nearly everything in the air.
Question 15 of 60 · Fluid Mechanics and Hydraulics · 12–18 items
Air flows through a rectangular HVAC duct with dimensions 2.0 m wide by 1.0 m high. Calculate the hydraulic diameter Dh to be used for Reynolds number and friction loss calculations.
Given
Duct width = 2.0 m
Duct height = 1.0 m
Find
Hydraulic diameter Dh (m)
Solution
Noncircular conduits use the hydraulic diameter in Re and Darcy–Weisbach (Fluid mechanics: noncircular conduits).
Dh = 4A/P = 4(2.0 m × 1.0 m)/(2 × 3.0 m) = 1.33 m
0.33 m is the hydraulic radius A/P; the hydraulic diameter is four times that.
5.10 kN is the log's weight, not the net upward force the tether resists.
7.33 kN leaves the unit weight of water out of the log's weight only (0.65 × 0.8 = 0.52) and subtracts that from the 7.85 kN buoyancy.
7.85 kN is the buoyant force alone; the log's weight offsets part of it.
12.9 kN adds buoyancy and weight; they act in opposite directions.
Question 17 of 60 · Thermodynamics · 3–5 items
A vertical plate (A = 2.00 m2) at 80 °C faces surroundings at 20 °C. The surface emissivity is ε = 0.85, and the convective coefficient is h = 12 W/(m2·K). Compute the total heat loss by radiation and convection.
Given
Plate area A = 2.00 m2
Surface temperature = 80 °C
Surroundings = 20 °C
Emissivity ε = 0.85
Convective coefficient h = 12 W/(m2·K)
Find
Total heat loss (kW)
Solution
Convection plus radiation to large surroundings, temperatures in kelvin for radiation (Heat transfer: radiation).
A closed system contains an ideal gas at 1.00 atm and 25.0 °C with a volume of 10.0 L. It is compressed reversibly and isothermally to 2.50 L. Calculate the work done by the gas. Use 1 L·atm = 101.325 J.
Given
Initial pressure P1 = 1.00 atm
Temperature = 25.0 °C
Initial volume V1 = 10.0 L
Final volume V2 = 2.50 L
1 L·atm = 101.325 J
Find
Work done by the gas (kJ)
Solution
Reversible isothermal work for an ideal gas (Thermodynamics: closed systems).
+1.40 kJ is the work done on the gas; work done by a gas being compressed is negative.
−0.760 kJ uses P1(V2 − V1), the constant-pressure work; an isothermal path needs the logarithm.
Question 19 of 60 · Surface Water Resources and Hydrology · 9–14 items
Segment
Length (m)
Slope (m/m)
Manning n
Depth (m)
Sheet flow
40
0.010
0.015
0.002
Shallow concentrated
150
0.020
0.030
0.050
Defined channel (rectangular)
1200
0.001
0.035
1.20
Compute the total time of concentration (Tc) in minutes for the three flow segments in the table. Use Manning's equation, v = (1/n)R2/3S1/2 (SI), and assume R ≈ depth for the sheet and shallow concentrated flow. The rectangular channel has a 3.0 m bottom width and vertical sides. Report Tc to one decimal place.
Given
Sheet flow: L = 40 m, S = 0.010, n = 0.015, depth 0.002 m
Shallow concentrated: L = 150 m, S = 0.020, n = 0.030, depth 0.050 m
Channel: L = 1,200 m, S = 0.001, n = 0.035, depth 1.20 m, width 3.0 m
Find
Time of concentration Tc (min)
Solution
Tc is the sum of segment travel times L/v, with v from Manning's equation (Hydrology: time of concentration).
sheet: v = (1/0.015)(0.002)2/3(0.010)1/2 = 0.106 m/s → 40/0.106 = 378 s = 6.3 min
shallow: v = (1/0.030)(0.050)2/3(0.020)1/2 = 0.640 m/s → 150/0.640 = 234 s = 3.9 min
channel: R = (3.0 × 1.20)/(3.0 + 2 × 1.20) = 0.667 m; v = 0.690 m/s → 1,200/0.690 = 1,740 s = 29.0 min
Tc = 6.3 + 3.9 + 29.0 = 39.2 min
26.3 minutes applies the US customary 1.49 factor in Manning's equation with SI lengths, so every velocity is 49% too high.
29.8 minutes uses the 1.20 m depth as R in the channel; for a 3.0 m wide channel R = A/P = 3.6/5.4 = 0.667 m.
32.9 minutes leaves out the 6.3-minute sheet-flow segment; Tc is the sum of all three travel times.
35.3 minutes leaves out the 3.9-minute shallow concentrated flow segment.
Travel times are 6.3 (sheet, v = 0.106 m/s) + 3.9 (shallow, v = 0.640 m/s) + 29.0 (channel, R = 0.667 m, v = 0.690 m/s) = 39.2 minutes.
Question 20 of 60 · Surface Water Resources and Hydrology · 9–14 items
Calculate the Muskingum routing coefficient C1 for a river reach with travel time K = 4 hours, weighting factor x = 0.2, and a routing time step Δt = 2 hours. The Muskingum equation is Ot+1 = C0 It+1 + C1 It + C2 Ot.
Given
Travel time K = 4 h
Weighting factor x = 0.2
Time step Δt = 2 h
Find
Muskingum coefficient C1
Solution
Muskingum coefficients share one denominator and sum to 1 (Hydrology: flood routing).
0.524 is C2 = (K − Kx − 0.5Δt)/D, the coefficient on the previous outflow Ot.
0.538 uses the full Δt instead of Δt/2 in both terms (2.8/5.2).
Question 21 of 60 · Surface Water Resources and Hydrology · 9–14 items
A hydrologist is modeling a river reach that consists of a deep main channel and wide, shallow floodplains. When flows exceed bankfull stage, the hydraulic properties change drastically. Which limitation makes the standard fixed-parameter Muskingum method unsuitable for this specific scenario?
Solution
Fixed K and x assume a constant wave celerity; overbank flow onto wide floodplains changes celerity with stage, so the parameters must vary (Hydrology: flood routing).
The opposite holds: Δt should not exceed about K (2Kx ≤ Δt ≤ 2K(1 − x)) to keep the coefficients positive.
Muskingum works on storage–flow relations and has no cross-section shape requirement.
Muskingum attenuates through its storage terms; with valid coefficients it is stable, and diffusion is not the floodplain problem.
When flow spreads onto wide, shallow floodplains the wave celerity drops sharply, so a single fixed K (and x) cannot describe both in-bank and overbank flow.
Storage is exactly what Muskingum represents, as prism storage KO plus wedge storage Kx(I − O).
Question 22 of 60 · Fluid Mechanics and Hydraulics · 12–18 items
A 300 mm diameter PVC pipe (E = 2.8 GPa) with a wall thickness of 15 mm carries water (K = 2.2 GPa). If the velocity drops instantaneously by 1.0 m/s, what is the approximate maximum pressure surge? Assume water density is 1,000 kg/m3.
Given
Pipe diameter D = 300 mm
Pipe modulus E = 2.8 GPa
Wall thickness e = 15 mm
Water bulk modulus K = 2.2 GPa
Velocity change ΔV = 1.0 m/s
Density ρ = 1,000 kg/m3
Find
Maximum pressure surge (kPa)
Solution
Wave celerity in an elastic pipe, then Joukowsky (Fluid mechanics: water hammer).
a = √{(2.2 × 109/1,000)/[1 + (2.2/2.8)(300/15)]} = 363 m/s
Δp = ρaΔV = 1,000 kg/m3 × 363 m/s × 1.0 m/s = 363 kPa
726 kPa doubles the Joukowsky rise; ρaΔV is already the full surge for an instantaneous change.
1,110 kPa leaves the D/e factor out of the wall term, so the PVC wall looks far stiffer than it is.
1,480 kPa uses the celerity in a rigid pipe, √(K/ρ); a PVC wall stretches and slows the wave.
1,670 kPa takes the celerity as √(E/ρ) using the pipe modulus instead of the water's bulk modulus.
Question 23 of 60 · Surface Water Resources and Hydrology · 9–14 items
A stream with a flow of 8.0 m3/s, DO of 8.5 mg/L, and BOD of 2.0 mg/L mixes with a wastewater discharge of 2.0 m3/s, DO of 1.0 mg/L, and BOD of 100 mg/L. The saturation DO is 10.0 mg/L. The deoxygenation rate (k1) is 0.2 day−1 and the reaeration rate (k2) is 0.4 day−1. What is the DO deficit 2.0 days downstream?
Given
Stream: Q = 8.0 m3/s, DO = 8.5 mg/L, BOD = 2.0 mg/L
Wastewater: Q = 2.0 m3/s, DO = 1.0 mg/L, BOD = 100 mg/L
Saturation DO = 10.0 mg/L
Deoxygenation rate k1 = 0.2 day−1
Reaeration rate k2 = 0.4 day−1
Travel time t = 2.0 days
Find
DO deficit (mg/L)
Solution
Mix first, then apply Streeter–Phelps (Surface water: DO sag).
Mixed DO 7.0 mg/L (Da = 3.0), mixed BOD 21.6 mg/L; D = (0.2 × 21.6/0.2)(e−0.4 − e−0.8) + 3.0e−0.8 = 4.77 + 1.35 = 6.1 mg/L.
7.8 mg/L adds the initial 3.0 mg/L deficit without letting it decay by e−k2t.
8.8 mg/L takes the initial deficit from the wastewater DO alone (10.0 − 1.0 = 9.0 mg/L) instead of the mixed 7.0 mg/L.
11.6 mg/L swaps k1 and k2 in the Streeter–Phelps equation.
23.4 mg/L uses the undiluted 100 mg/L wastewater BOD instead of the mixed 21.6 mg/L.
Question 24 of 60 · Groundwater, Soils, and Sediments · 8–12 items
A homogeneous horizontal aquifer has hydraulic conductivity K = 10 m/day, hydraulic gradient i = 0.002, and effective porosity ne = 0.25. A sorbing contaminant with retardation factor R = 5 originates 500 m upgradient. Estimate the travel time to the receptor (days).
Given
Hydraulic conductivity K = 10 m/day
Hydraulic gradient i = 0.002
Effective porosity ne = 0.25
Retardation factor R = 5
Travel distance L = 500 m
Find
Contaminant travel time (days)
Solution
Retarded travel time (Groundwater: contaminant transport).
125,000 days applies R to the Darcy flux instead of the seepage velocity.
Question 25 of 60 · Fundamental Principles · 7–11 items
A Continuous Stirred-Tank Reactor (CSTR) is designed to treat wastewater with a flow rate of 500 m3/d. The pollutant decays via first-order kinetics with a rate constant k = 4.0 d−1. What reactor volume is required to achieve 80% removal efficiency?
Given
Flow Q = 500 m3/d
Rate constant k = 4.0 d−1
Required removal = 80 %
Find
Required CSTR volume (m3)
Solution
CSTR with first-order decay (Fundamental principles: reactor design).
125 m3 is Q/k alone; the removal target, (Cin/Cout − 1) = 4, never enters.
201 m3 uses the plug-flow equation τ = ln(5)/k = 0.40 d; the reactor is a CSTR.
Cout/Cin = 0.20 = 1/(1 + 4.0τ) gives τ = 1.0 d, so V = 500 m3/d × 1.0 d = 500 m3.
625 m3 uses τ = (Cin/Cout)/k = 1.25 d and forgets to subtract 1.
2,000 m3 uses kτ = 4 as if τ were 4 days, without dividing by k = 4.0 d−1.
Question 26 of 60 · Water and Wastewater · 12–18 items
A wastewater treatment plant operates a centrifuge for sludge dewatering. The unit receives a feed sludge flow of 25 m3/hr with a solids concentration of 3.0% (assume specific gravity of 1.0). The centrifuge operates with a solids capture efficiency of 95% and produces a dewatered cake with 22% solids. Calculate the mass loading of dry solids returning to the headworks in the centrate stream.
Given
Feed sludge flow = 25 m3/hr
Feed solids = 3.0%
Specific gravity = 1.0
Solids capture = 95%
Cake solids = 22%
Find
Dry solids in centrate (kg/hr)
Solution
Solids capture splits the feed solids between cake and centrate (Water and wastewater: sludge dewatering).
Question 27 of 60 · Water and Wastewater · 12–18 items
An anaerobic digester treats 100 m3/d of primary sludge with 5.0% total solids (TS), of which 75% are volatile solids (VS). The digester achieves 60% VS destruction. The specific methane yield is 0.50 m3 CH4 per kg VS destroyed. If the lower heating value of methane is 35.8 MJ m−3 and the cogeneration system has an electrical efficiency of 35%, estimate the daily electricity generation.
Given
Sludge flow = 100 m3/d
Total solids = 5.0 %
Volatile fraction = 75 %
VS destruction = 60 %
Methane yield = 0.50 m3 CH4/kg VS destroyed
Methane LHV = 35.8 MJ/m3
Electrical efficiency = 35 %
Find
Daily electricity generation (kWh/d)
Solution
VS fed → VS destroyed → methane volume → electrical energy (Water and wastewater: anaerobic digestion).
6,526 kWh/d uses all 3,750 kg/d of VS fed instead of the 60% destroyed.
11,188 kWh/d is the full heating value of the methane (100% conversion), before the 35% electrical efficiency.
Question 28 of 60 · Water and Wastewater · 12–18 items
A rainwater harvesting system utilizes a catchment area of 200 m2 with a runoff coefficient of 0.90. During a storm event with a precipitation depth of 50 mm, a first-flush diverter removes the initial 200 L of runoff before water enters the storage tank. What is the net volume of water captured?
Given
Catchment area = 200 m2
Runoff coefficient = 0.90
Precipitation depth = 50 mm
First-flush diversion = 200 L
Find
Net captured volume (L)
Solution
Captured volume = area × depth × runoff coefficient − first flush (Water resources: rainwater harvesting).
V = 200 m2 × 0.050 m × 0.90 = 9.0 m3 = 9,000 L
Vnet = 9,000 − 200 = 8,800 L
Runoff = 200 m2 × 0.050 m × 0.90 = 9.0 m3 = 9,000 L; minus the 200 L first flush = 8,800 L.
9,000 L is the runoff before the 200 L first-flush diversion.
9,200 L adds the 200 L first flush to the runoff instead of diverting it.
9,800 L subtracts the first flush but uses a runoff coefficient of 1.0.
10,000 L is the rain falling on the roof, with no runoff coefficient and no first flush.
Question 29 of 60 · Water and Wastewater · 12–18 items
A plant must reduce influent NO3-N from 21.0 mg/L to 1.0 mg/L by denitrification. Readily biodegradable internal COD available is 30.0 mg/L. The theoretical COD demand for nitrate reduction is 2.86 kg COD per kg NO3-N reduced. What supplemental COD (mg/L) is required?
Given
Influent NO3-N = 21.0 mg/L
Effluent NO3-N = 1.0 mg/L
Available internal COD = 30.0 mg/L
COD demand = 2.86 kg COD/kg NO3-N
Find
Supplemental COD (mg/L)
Solution
Denitrification COD demand minus the internal COD (Water and wastewater: nitrogen removal).
CODreq = (21.0 − 1.0) mg N/L × 2.86 = 57.2 mg/L
CODsupp = 57.2 − 30.0 = 27.2 mg/L
87.2 mg/L adds the available COD to the demand instead of subtracting it.
61.4 mg/L uses 4.57, the oxygen demand of nitrification, instead of 2.86 g COD per g NO3-N for denitrification.
57.2 mg/L is the total COD demand; the 30.0 mg/L already available reduces what must be added.
30.1 mg/L uses all 21.0 mg/L of nitrate; the effluent may keep 1.0 mg/L.
Question 30 of 60 · Water and Wastewater · 12–18 items
An MBR must treat an average dry weather flow of 6,000 m3/day. The design uses a sustainable flux Js = 20 L/(m2·hr) and a short-term maximum flux Jmax = 40 L/(m2·hr) for peak flows up to 3× average. Determine the membrane area (m2) required by the governing condition.
Given
Average flow Qavg = 6000 m3/day
Sustainable flux Js = 20 L/(m2·hr)
Maximum flux Jmax = 40 L/(m2·hr)
Peak flow = 3 × average
Find
Governing membrane area (m2)
Solution
Check each flow condition against its allowed flux; the larger area governs (Water and wastewater: membranes).
6,250 m2 sizes the average flow at the short-term maximum flux; Jmax is allowed only during peaks.
12,500 m2 covers average flow at the sustainable flux, but the 3× peak at 40 LMH needs more.
Average: 250,000 L/hr ÷ 20 = 12,500 m2; peak: 750,000 L/hr ÷ 40 = 18,750 m2. The peak case governs.
37,500 m2 holds the peak flow to the sustainable flux; peaks may run at Jmax.
300,000 m2 divides L/day by a flux in L/(m2·hr) without converting days to hours.
Question 31 of 60 · Water and Wastewater · 12–18 items
A plant with flow 5000 m3/day has influent total phosphorus 6.0 mg/L (30 kg P/day). Net waste activated sludge production is 500 kg VSS/day. Biomass P content is 0.02 kg P per kg VSS. Calculate the daily phosphorus that must be removed chemically (kg P/day) after biological uptake.
Given
Flow = 5000 m3/day
Influent TP = 6.0 mg/L
Net WAS production = 500 kg VSS/day
Biomass P content = 0.02 kg P/kg VSS
Find
Phosphorus to remove chemically (kg P/day)
Solution
Phosphorus balance (Water and wastewater: nutrient removal).
30 kg P/day is the whole influent load, ignoring the 10 kg/day taken up by biomass.
Question 32 of 60 · Groundwater, Soils, and Sediments · 8–12 items
A contaminant plume is migrating through a confined aquifer composed of fine sand. The hydraulic conductivity (K) is 5.0 × 10−4 cm/s, the hydraulic gradient (i) is 0.01, and the effective porosity (ne) is 0.25. Based on the average linear velocity, how much time is required for the center of mass of a conservative tracer to travel 1.5 km?
Given
Hydraulic conductivity K = 5.0 × 10−4 cm/s
Hydraulic gradient i = 0.01
Effective porosity ne = 0.25
Travel distance = 1.5 km
Find
Travel time of the center of mass (yr)
Solution
Travel time at the average linear velocity (Groundwater: Darcy's law).
q = Ki = 5.0 × 10−6 cm/s, v = q/ne = 2.0 × 10−5 cm/s = 6.31 m/yr, t = 1,500/6.31 = 238 yr.
951 yr uses the Darcy flux as the travel velocity, ignoring the effective porosity.
2,380 yr converts cm/s to m/s with 1/1,000 instead of 1/100, making the velocity ten times too small.
3,800 yr multiplies the Darcy flux by ne instead of dividing.
14,300 yr converts with 525,600 s per year (the number of minutes); a year has 3.16 × 107 s.
Question 33 of 60 · Energy and Environment · 4–6 items
An 800-MW nuclear power plant operates with a heat rate of 10,400 Btu/kWh. What is the approximate rate of waste heat rejection to the environment in MW?
Given
Plant output = 800 MW
Heat rate = 10,400 Btu/kWh
Find
Waste heat rejected (MW)
Solution
Heat rate gives the thermal input; waste heat = input − electrical output (Energy and environment: power plants).
Qin = 800 MW × 10,400/3,412 = 2,440 MW
Qrejected = 2,440 − 800 = 1,640 MW
538 MW multiplies the electrical output by (1 − η); the rejected heat is the input minus the output.
800 MW is the electrical output, not the heat rejected.
2,440 MW is the total thermal input; 800 MW of it leaves as electricity.
3,240 MW adds the output to the input instead of subtracting it.
Question 34 of 60 · Energy and Environment · 4–6 items
A proposed 100 MW wind farm has an expected service life of 20 years and a capacity factor of 30%. The embodied (construction) carbon intensity is 1,500 kg CO2e/kW installed. Operational maintenance emissions are estimated at 5 g CO2e/kWh. What is the total estimated lifetime carbon footprint of the facility in metric tons CO2e?
150,000 metric tons is the embodied (construction) share alone.
176,280,000 is the footprint in kilograms labeled as metric tons.
26,280 metric tons is the operational share alone; construction adds 150,000 metric tons.
237,600 metric tons ignores the 30% capacity factor, as if the turbines ran at full output all year.
Question 35 of 60 · Solid and Hazardous Waste · 7–11 items
A landfill cell has a covered area of 40,000 m2. Using the monthly climate data below, compute the annual volume of water that percolates to the liner with a month-by-month balance: infiltration = max(0, P − ET) × (1 − runoff coefficient), then convert millimeters to cubic meters.
Month
Precip (mm)
ET (mm)
Runoff coeff
Jan
40
10
0.10
Feb
35
15
0.10
Mar
45
25
0.12
Apr
50
40
0.15
May
60
50
0.20
Jun
70
60
0.25
Jul
80
70
0.30
Aug
90
70
0.30
Sep
70
50
0.25
Oct
40
30
0.15
Nov
40
15
0.12
Dec
40
5
0.10
Given
Cell area = 40,000 m2
Monthly P, ET and runoff coefficient as tabulated
Find
Annual percolation to the liner (m3/yr)
Solution
Monthly water balance on the cover, summed over the year (Solid waste: landfill leachate).
7,380,000 multiplies millimeters by square meters without converting mm to m.
73,800 m3/yr divides millimeters by 100 instead of 1,000 when converting to meters.
21,100 m3/yr removes runoff but never subtracts evapotranspiration.
8,800 m3/yr uses the full surplus P − ET (220 mm) and ignores the runoff coefficient.
Monthly infiltration max(0, P − ET)(1 − C) sums to 184.6 mm; 0.1846 m × 40,000 m2 = 7,380 m3/yr.
Question 36 of 60 · Solid and Hazardous Waste · 7–11 items
A geonet leachate drainage layer has a transmissivity θ = 2.5 × 10−4 m2/s on a slope of 0.5% (i = 0.005); a combined clogging and creep reduction factor of 0.60 applies. The design leachate impingement rate is 2.0 mm/day (0.0020 m3/day per m2 of liner). If all leachate falling between two pipes drains downslope to one pipe, so the flow per meter of pipe equals the impingement rate times the pipe spacing, what is the maximum pipe spacing? Use a drainage capacity per unit width of q = θ·i.
Given
Geonet transmissivity θ = 2.5 × 10−4 m2/s
Slope i = 0.005
Reduction factor = 0.60
Leachate impingement rate = 2.0 mm/day
Find
Maximum pipe spacing S (m)
Solution
Allowable flow per unit width must carry the leachate collected over the spacing (Solid waste: leachate collection).
Question 37 of 60 · Solid and Hazardous Waste · 7–11 items
A composite liner has four defects, each with an area of 1 cm2. For one defect, kd = 1.0 × 10−4 m/s, h = 1.0 m, and Leff = 0.005 m, with Qdefect ≈ kdAd(h/Leff). The total annual infiltration reaching the liner is 20,000 m3/yr. Estimate the leachate collection efficiency.
Given
Number of defects = 4
Defect area Ad = 1 cm2 each
kd = 1.0 × 10−4 m/s
Head h = 1.0 m
Leff = 0.005 m
Annual infiltration = 20,000 m3/yr
Find
Leachate collection efficiency (%)
Solution
Leakage through liner defects by Darcy's law, compared with the water reaching the liner (Solid waste: liner performance).
6.0 is the first divided difference f[x1, x2], not the second.
4.0 is the difference of the first divided differences (6 − 2) before dividing by x2 − x0 = 2.
f[x0, x1] = 2, f[x1, x2] = 6, so f[x0, x1, x2] = (6 − 2)/(2 − 0) = 2.0; the second divided difference is the x2 coefficient.
0.0 is the coefficient of the x term once the Newton polynomial 2x + 2x(x − 1) is expanded to 2x2, not the x2 coefficient.
Question 39 of 60 · Probability and Statistics · 4–6 items
A remediation pilot study reports contaminant concentrations (mg/L) at five wells before and after treatment. The paired data (Before, After) are: (10, 9), (12, 14), (15, 11), (20, 15), and (18, 18). Calculate the Wilcoxon signed-rank test statistic, W, using differences defined as d = Before - After and summing the signed ranks.
Wilcoxon signed-rank: drop zero differences, rank |d|, attach signs, sum (Probability and statistics: nonparametric tests).
d = 1, −2, 4, 5 (one zero dropped) → ranks 1, 2, 3, 4
W = +1 − 2 + 3 + 4 = 6
−6 defines the differences as After − Before; the stem fixes d = Before − After.
2 is W−, the sum of the negative ranks only.
d = +1, −2, +4, +5 (the zero is dropped); ranks of |d| are 1, 2, 3, 4; signed sum = 1 − 2 + 3 + 4 = 6.
8 is W+, the sum of the positive ranks only, not the signed sum.
10 adds all ranks and ignores their signs.
Question 40 of 60 · Engineering Economics · 5–8 items
A project requires a $10,000 investment in depreciable assets with a 4-year class life and no salvage value. The project generates $5,000 in annual real pre-tax savings. The real minimum attractive rate of return (MARR) is 5%, inflation is 5%, and the tax rate is 40%. Using straight-line depreciation, calculate the after-tax present worth.
Given
Investment = $10,000
Class life = 4 years, no salvage, straight-line
Annual real pre-tax savings = $5,000
Real MARR = 5 %
Inflation = 5 %
Tax rate = 40 %
Find
After-tax present worth ($)
Solution
Discount real cash flows at the real rate and nominal ones (the depreciation shield) at the nominal rate (Engineering economics: inflation and taxes).
$638 leaves out the depreciation tax shield entirely.
Question 41 of 60 · Surface Water Resources and Hydrology · 9–14 items
A 200-acre watershed has the hydrologic characteristics listed below. Using the SCS Curve Number method and the standard SCS unit hydrograph peak discharge equation (qp = (484 A Q)/Tp), calculate the peak flow of the direct runoff hydrograph for the design storm.
Parameter
Value
Watershed Area
200 acres
Curve Number (CN)
80
Time to Peak (Tp)
2.0 hours
Design Storm Rainfall (P)
5.0 inches
Given
Watershed area = 200 acres
Curve number CN = 80
Time to peak Tp = 2.0 h
Design rainfall P = 5.0 in
Find
Peak direct runoff qp (cfs)
Solution
Curve-number runoff depth, then the SCS unit-hydrograph peak with A in mi2 (Hydrology: NRCS method).
S = 1000/80 − 10 = 2.5 in; Ia = 0.5 in; Q = (5.0 − 0.5)2/(5.0 − 0.5 + 2.5) = 2.89 in
qp = 484AQ/Tp = 484 × 0.3125 mi2 × 2.89 in / 2.0 h = 219 cfs
S = 2.5 in, Ia = 0.5 in, Q = 4.52/7.0 = 2.89 in; qp = 484 × (200/640) mi2 × 2.89 in / 2.0 h = 219 cfs.
252 cfs drops the initial abstraction, Q = P2/(P + S) = 3.33 in.
340 cfs uses P − Ia = 4.5 in as the runoff depth and leaves out the retention S.
378 cfs uses all 5.0 in of rainfall as runoff.
567 cfs converts 200 acres to km2 (0.809) instead of mi2 (0.3125); the 484 factor needs square miles.
Question 42 of 60 · Surface Water Resources and Hydrology · 9–14 items
A lake with a volume of 1.0 × 107 m3 receives an external phosphorus load of 2,000 kg/yr. The outflow rate is 5.0 × 106 m3 yr−1 and the phosphorus settling rate constant is 0.40 yr−1. If the steady-state phosphorus concentration is 0.30 mg/L, what is the estimated internal phosphorus loading rate from sediment release?
Given
Lake volume V = 1.0 × 107 m3
External P load = 2,000 kg/yr
Outflow Q = 5.0 × 106 m3/yr
Settling rate k = 0.40 yr−1
Steady-state P = 0.30 mg/L
Find
Internal P loading (kg/yr)
Solution
Steady-state lake mass balance: inputs = outflow + settling (Surface water: lake models).
Question 43 of 60 · Groundwater, Soils, and Sediments · 8–12 items
A well fully penetrates a confined aquifer and is pumped at a steady rate of 2.5 m3/min. Drawdowns observed at distances of 30 m and 120 m from the well are 4.5 m and 1.2 m, respectively. The transmissivity of the aquifer is most nearly:
Given
Pumping rate Q = 2.5 m3/min
Drawdown s1 = 4.5 m at r1 = 30 m
Drawdown s2 = 1.2 m at r2 = 120 m
Find
Transmissivity T (m2/day)
Solution
Thiem equation with two observation wells (Groundwater: well hydraulics).
105 m2/day uses log10(r2/r1) instead of the natural log.
139 m2/day adds the drawdowns; the Thiem equation uses their difference, s1 − s2 = 3.3 m.
Question 44 of 60 · Air Quality and Control · 8–12 items
A facility with 4,000 m3/h of supply air (15% outdoor air) generates 2,100 mg/h of particulates. The outdoor concentration is 0.20 mg/m3. Filtration treats only the recirculated air. To keep the indoor concentration at or below 1.0 mg/m3, select the most economical filter.
Model
Efficiency
Installed Cost
F-25
25%
$120
F-30
30%
$150
F-45
45%
$220
F-65
65%
$290
F-85
85%
$350
Given
Supply air Q = 4,000 m3/h
Outdoor air fraction = 15%
Particulate generation G = 2,100 mg/h
Outdoor concentration Cout = 0.20 mg/m3
Indoor limit Cin = 1.0 mg/m3
Find
Most economical filter
Solution
Steady-state room balance with filtered recirculation (Air quality: indoor air).
Balance: 2,100 + 600 × 0.20 + 3,400(1 − η)(1.0) = 4,000 × 1.0 gives η = 47.6%; F-65 is the cheapest filter above that.
F-30 removes only 30% of the recirculated load; the balance needs η ≥ 47.6%, so the room would stay above 1.0 mg/m3.
F-85 also meets the limit, but it costs $60 more than F-65, which is already sufficient.
F-25 (25%) is far below the 47.6% the recirculated air must have removed.
F-45 (45%) is just short of the 47.6% needed; it would pass only if the 0.20 mg/m3 brought in with outdoor air were ignored.
Question 45 of 60 · Water and Wastewater · 12–18 items
A facility needs 15 m3/d of process water (TDS ≤ 80 mg/L) produced by blending rainwater membrane permeate (TDS 10 mg/L, 75% recovery) with well water (TDS 350 mg/L). Annual rainfall is 1,100 mm (C = 0.85). The required roof catchment area is most nearly:
Given
Process water demand = 15 m3/d
Target TDS ≤ 80 mg/L
Permeate TDS = 10 mg/L
Membrane recovery = 75 %
Well water TDS = 350 mg/L
Annual rainfall = 1,100 mm
Runoff coefficient C = 0.85
Find
Required roof catchment area (m2)
Solution
Blend mass balance → permeate → membrane feed → roof area (Water resources: alternative supplies).
15 × 80 = 10Qp + 350(15 − Qp) → Qp = 11.9 m3/d
Qfeed = 11.9/0.75 = 15.9 m3/d = 5,800 m3/yr
A = 5,800 m3/yr ÷ (1.1 m/yr × 0.85) = 6,200 m2
3,490 m2 multiplies the permeate by the 75% recovery instead of dividing; the membrane needs more feed than it delivers.
4,650 m2 sizes the roof for the permeate flow and ignores the 25% of feed lost to concentrate.
5,270 m2 leaves out the runoff coefficient C = 0.85.
7,810 m2 makes all 15 m3/d from permeate; blending with well water lets part of the demand bypass the membrane.
Question 46 of 60 · Air Quality and Control · 8–12 items
Condition
Stability
σy (m)
σz (m)
Plume Rise Δh (m)
Strong Insolation
A
110
80
30
Overcast/Neutral
D
40
20
15
Clear Night
F
20
10
5
A facility emits 80 g/s of sulfur dioxide from a stack with physical height h = 40 m. The wind speed is 3 m/s at stack height. Using the dispersion parameters in the table, calculate the centerline ground-level concentration at x = 500 m for the worst-case meteorological condition.
852 µg/m3 uses the physical stack height (40 m) and ignores the 30 m plume rise.
242 µg/m3 is the neutral class D result, which is lower than class A at this distance.
329 µg/m3 uses Q/(2πuσyσz) at ground level and drops the ground-reflection term, which doubles the concentration at z = 0.
1.70 µg/m3 is the stable class F result; at 500 m the narrow plume has not yet reached the ground.
Class A, H = 40 + 30 = 70 m: C = [80/(π × 3 × 110 × 80)] × e−702/(2 × 802) = 9.65 × 10−4 × 0.682 g/m3 = 658 µg/m3.
Question 47 of 60 · Air Quality and Control · 8–12 items
An electrostatic precipitator (ESP) is designed to treat a gas flow of 60 m3/s with an effective drift velocity of 0.12 m s−1. The inlet particulate loading is 5,000 kg/day, and the outlet emissions must not exceed 50 kg/day. The required collection plate area is most nearly:
Given
Gas flow Q = 60 m3/s
Effective drift velocity w = 0.12 m/s
Inlet particulate loading = 5,000 kg/day
Allowed outlet emissions = 50 kg/day
Find
Required collection plate area A (m2)
Solution
Deutsch–Anderson solved for plate area (Air quality: electrostatic precipitators).
495 m2 multiplies Q/w by the efficiency 0.99 instead of by −ln(1 − η).
1,000 m2 uses log10(1/0.01) = 2 instead of the natural log, 4.61.
1,150 m2 halves the area as if each plate counted twice; A in the Deutsch equation is the total collecting area.
Required η = (5,000 − 50)/5,000 = 0.99; A = −(Q/w)ln(1 − η) = (60/0.12) × 4.61 = 2,300 m2.
Question 48 of 60 · Air Quality and Control · 8–12 items
Parameter
Measured Value
NOx concentration (dry)
45 ppmvd
Oxygen concentration (dry)
3.0%
F-factor (dry)
8,710 dscf/MMBtu
NO2 molecular weight
46.01 lb/lbmol
A natural-gas-fired boiler stack test measures NOx at 45 ppmvd (parts per million by volume, dry) at an oxygen concentration of 3.0% (dry). Using EPA Method 19 and the data provided, calculate the NOx emission rate in lb/MMBtu (as NO2). The standard molar volume is 385.3 dscf/lbmol at 68 °F.
When using finite difference methods to numerically approximate the derivative of a thermodynamic property, decreasing the step size (h) initially improves accuracy. However, if h becomes extremely small, the total computed error begins to increase. Which of the following best explains this phenomenon?
Solution
Total error = truncation (∝ hp) + round-off (∝ εmachine/h); the sum has a minimum at an optimal h (Mathematics: numerical methods).
No matrix is inverted in a difference quotient, and a well-conditioned problem has a condition number near 1, not 0.
A finite-difference derivative is a direct formula, not an optimization, so there is no local minimum to get trapped in.
Changing h does not turn an explicit difference formula into an implicit one.
Truncation error keeps shrinking with h (as h or h2); it does not reverse at small steps.
Truncation error falls as h shrinks, but the difference f(x + h) − f(x) loses significant digits; that round-off error grows roughly as 1/h and eventually dominates.
Question 50 of 60 · Environmental Chemistry · 7–11 items
In a batch test with excess substrate, a bacterial culture grows exponentially at a specific growth rate μ = 0.30 h−1. How long does the biomass take to rise from 50 mg/L to 400 mg/L?
Given
Initial biomass X0 = 50 mg/L
Final biomass X = 400 mg/L
Specific growth rate μ = 0.30 h−1
Find
Time t (h)
Solution
Exponential (unlimited) growth in batch (Environmental chemistry: microbiology).
t = ln(X/X0)/μ = ln(8)/0.30 h−1 = 6.93 h
2.31 h is one doubling time (ln 2/μ); the biomass must double three times to go from 50 to 400 mg/L.
3.01 h uses log10(8) instead of the natural log.
X = X0eμt → t = ln(400/50)/0.30 = 2.079/0.30 = 6.93 h (three doublings of 2.31 h).
23.3 h assumes linear growth at the initial rate (350/(0.30 × 50)).
26.7 h divides the ratio 8 by μ instead of its natural log.
Question 51 of 60 · Environmental Chemistry · 7–11 items
A wastewater contains 300 mg/L of glucose (C6H12O6, molar mass 180 g/mol). Assuming complete oxidation C6H12O6 + 6O2 → 6CO2 + 6H2O, what is its theoretical oxygen demand?
Given
Glucose = 300 mg/L
Molar mass of glucose = 180 g/mol
Molar mass of O2 = 32 g/mol
Find
Theoretical oxygen demand (mg/L)
Solution
Oxygen required by stoichiometry (Environmental chemistry: oxygen demand).
ThOD = C × (6 × 32 g O2)/(180 g glucose) = 300 × 1.067 = 320 mg/L
53.3 mg/L uses one O2 per glucose; the balanced equation needs six.
160 mg/L counts 6 oxygen atoms (96 g) per mole of glucose instead of 6 O2 molecules (192 g).
281 mg/L inverts the mass ratio (180/192 instead of 192/180).
300 mg/L assumes one gram of oxygen per gram of glucose.
Question 52 of 60 · Solid and Hazardous Waste · 7–11 items
A composting facility blends dewatered biosolids (dry basis: 30% C, 5.0% N) with sawdust (dry basis: 50% C, 0.20% N). How many kilograms of dry sawdust are needed per kilogram of dry biosolids to reach a blended C:N ratio of 25?
Given
Biosolids: C = 30%, N = 5.0% (dry)
Sawdust: C = 50%, N = 0.20% (dry)
Target C:N = 25
Find
Dry sawdust per kg dry biosolids (kg)
Solution
Blend by mass balance on carbon and nitrogen, then set the ratio (Solid waste: composting).
C:N = (0.30 + 0.50x)/(0.050 + 0.0020x) = 25
0.30 + 0.50x = 1.25 + 0.050x → x = 2.11 kg sawdust per kg biosolids
2.78 kg ignores the carbon already supplied by the biosolids.
1.90 kg ignores the small nitrogen content of the sawdust.
0.47 kg is the inverse, kilograms of biosolids per kilogram of sawdust.
0.084 kg averages the C:N ratios themselves (6 and 250); ratios do not blend linearly by mass.
Question 53 of 60 · Fundamental Principles · 7–11 items
Stream A (Q = 2 MGD, BOD = 50 mg/L) mixes with Stream B (Q = 3 MGD, BOD = 10 mg/L). The combined flow enters a plug flow reactor with a volume of 0.5 MG and a decay coefficient k = 2.5 d−1. Calculate the effluent BOD concentration.
Given
Stream A: Q = 2 MGD, BOD = 50 mg/L
Stream B: Q = 3 MGD, BOD = 10 mg/L
Reactor volume V = 0.5 MG
Decay coefficient k = 2.5 d−1
Find
Effluent BOD (mg/L)
Solution
Flow-weighted blend, then first-order decay in plug flow (Fundamental principles: mass balance and reactors).
23.4 mg/L averages the two BOD values (30 mg/L) without weighting by flow before applying the decay.
26.0 mg/L is the blended influent; the reactor still removes BOD.
Question 54 of 60 · Fluid Mechanics and Hydraulics · 12–18 items
A rectangular barge is 20 m long and 12 m wide and floats with a draft of 2.5 m in seawater. Its center of gravity G is 3.0 m above the keel. Calculate the metacentric height GM for rolling (about the long axis).
Given
Length L = 20 m
Width W = 12 m
Draft = 2.5 m
KG = 3.0 m
Find
Metacentric height GM (m)
Solution
Initial stability of a floating body: GM = KB + BM − KG with BM = I/V about the longitudinal axis (Fluid mechanics: buoyancy and stability).
KB = 2.5/2 = 1.25 m; I = LW3/12 = 20 × 123/12 = 2,880 m4; V = 20 × 12 × 2.5 = 600 m3
BM = 2,880/600 = 4.80 m; GM = 1.25 + 4.80 − 3.0 = 3.05 m
1.80 m drops KB and takes GM = BM − KG = 4.80 − 3.0.
4.30 m puts the center of buoyancy at the full draft (2.5 m) instead of half of it.
6.05 m is KM = KB + BM, the height of the metacenter; KG = 3.0 m must still be subtracted.
11.6 m takes the moment of inertia about the wrong axis, I = W·L3/12, which governs pitch, not roll.
Question 55 of 60 · Fluid Mechanics and Hydraulics · 12–18 items
A pressure gauge on a process tank at a high-altitude site reads a vacuum of 22 kPa. The local atmospheric pressure is 94 kPa. What is the absolute pressure in the tank?
Given
Gauge reading = 22 kPa vacuum
Local atmospheric pressure = 94 kPa
Find
Absolute pressure (kPa)
Solution
Absolute pressure = local atmospheric + gauge, with a vacuum counted as negative gauge (Fluid mechanics: pressure).
A vacuum gauge reads below local atmospheric, so Pabs = Patm − Pvac = 94 − 22 = 72 kPa.
79 kPa subtracts the vacuum from standard atmospheric pressure (101.3 kPa) instead of the local 94 kPa.
94 kPa is the local atmospheric pressure; it ignores the 22 kPa vacuum.
116 kPa adds the vacuum to the atmospheric pressure; a vacuum is a negative gauge pressure.
123 kPa adds the vacuum to standard atmospheric pressure, two errors at once.
Question 56 of 60 · Fluid Mechanics and Hydraulics · 12–18 items
A hydraulic jump forms in a wide rectangular channel carrying a unit discharge q = 4.00 m2/s. The depth just upstream of the jump is 0.60 m. Using g = 9.81 m/s2, compute the energy loss across the jump.
Given
Unit discharge q = 4.00 m2/s
Upstream depth y1 = 0.60 m
g = 9.81 m/s2
Find
Energy loss across the jump ΔE (m)
Solution
Sequent depth from the momentum equation, then the head loss across the jump (Hydraulics: hydraulic jump).
1.451 m is the rise in depth, y2 − y1, not the energy lost.
2.482 m leaves the 4 out of ΔE = (y2 − y1)3/(4y1y2).
2.865 m is the specific energy before the jump, E1 = y1 + q2/(2gy12), not the loss.
Question 57 of 60 · Groundwater, Soils, and Sediments · 8–12 items
An in-situ chemical oxidation (ISCO) pilot test treats a saturated aquifer volume of 200 m3. The aquifer has a porosity of 0.25, a soil bulk density of 1,600 kg m−3, and a Natural Oxidant Demand (NOD) of 1.5 g/kg. The groundwater contains 20 mg/L of TCE. The stoichiometric oxidant demand for TCE is 3.0 g oxidant per 1.0 g TCE. What is the minimum total mass of oxidant required to satisfy both the contaminant and soil demands?
Given
Treated aquifer volume = 200 m3
Porosity n = 0.25
Bulk density = 1,600 kg/m3
Natural oxidant demand = 1.5 g/kg
TCE in groundwater = 20 mg/L
Stoichiometric demand = 3.0 g oxidant/g TCE
Find
Minimum total oxidant mass (kg)
Solution
Oxidant dose = contaminant stoichiometric demand + natural oxidant demand of the soil (Groundwater: remediation).
TCE = 200 m3 × 0.25 × 20 g/m3 = 1.0 kg → 3.0 kg oxidant
NOD = 200 m3 × 1,600 kg/m3 × 1.5 g/kg = 480 kg
total = 483 kg
3 kg covers only the TCE (1.0 kg × 3.0); the soil's natural oxidant demand is far larger.
363 kg multiplies the bulk density by the solids fraction (1 − 0.25); bulk density already refers to the whole soil volume.
480 kg is the natural oxidant demand alone, without the 3 kg for TCE.
Pore water = 200 × 0.25 = 50 m3, TCE = 50,000 L × 20 mg/L = 1.0 kg → 3.0 kg oxidant; soil = 200 × 1,600 = 320,000 kg × 1.5 g/kg = 480 kg; total 483 kg.
492 kg puts 20 mg/L of TCE in the whole 200 m3 instead of only the 50 m3 of pore water.
Question 58 of 60 · Groundwater, Soils, and Sediments · 8–12 items
A Permeable Reactive Barrier (PRB) is designed to treat groundwater with a seepage velocity of 0.2 m/day. The influent concentration is 100 mg/L and the regulatory limit is 1.0 mg/L. If the reaction follows first-order kinetics with a rate constant k = 0.5 day−1, what is the minimum required thickness of the barrier?
Given
Seepage velocity v = 0.2 m/day
Influent concentration = 100 mg/L
Regulatory limit = 1.0 mg/L
Rate constant k = 0.5 day−1
Find
Minimum barrier thickness (m)
Solution
Residence time for first-order decay to the limit, then thickness = seepage velocity × time (Groundwater: permeable reactive barriers).
t = ln(C0/C)/k = ln(100/1.0)/0.5 day−1 = 9.21 days
L = v·t = 0.2 m/day × 9.21 days = 1.84 m
0.396 m treats the decay as linear: (1 − 0.01)/0.5 days × 0.2 m/day.
0.800 m uses log10(100) = 2 instead of ln(100) = 4.61.
Required residence time t = ln(100/1.0)/0.5 = 9.21 days; thickness = v·t = 0.2 × 9.21 = 1.84 m.
9.21 is the residence time in days; it still has to be multiplied by the 0.2 m/day velocity.
46.1 m divides the residence time by the velocity instead of multiplying.
Question 59 of 60 · Air Quality and Control · 8–12 items
A continuous emission monitoring system measures NOx at 36 ppmv on a wet basis. The stack gas contains 10% moisture by volume, and the regulatory limit is specified on a dry basis. What is the dry-basis NOx concentration?
Given
Wet-basis NOx = 36 ppmv
Moisture = 10% by volume
Find
Dry-basis NOx (ppmv)
Solution
Wet-to-dry correction removes the water vapor from the gas volume (Air quality: stack measurements).
Cdry = Cwet/(1 − Bws) = 36/(1 − 0.10) = 40.0 ppmv
32.4 ppmv multiplies by the dry fraction (36 × 0.90) instead of dividing.
32.7 ppmv divides by 1 + 0.10 instead of 1 − 0.10.
36.0 ppmv leaves the reading on the wet basis.
39.6 ppmv multiplies by 1.10 instead of dividing by 0.90.
Removing the 10% water vapor leaves 90% of the gas: Cdry = 36/(1 − 0.10) = 40.0 ppmv.
Question 60 of 60 · Energy and Environment · 4–6 items
Water use data for two power plants is provided below:
Plant A: Withdrawal = 40,000 gal/MWh; Consumption = 300 gal/MWh
Plant B: Withdrawal = 800 gal/MWh; Consumption = 600 gal/MWh
Based on these ratios, which statement correctly identifies the cooling system types?
Solution
Once-through: high withdrawal, low consumption; recirculating (wet tower): low withdrawal, most of it evaporated (Energy and environment: water use).
This reverses them: a recirculating tower would not need 40,000 gal/MWh of withdrawal.
Once-through systems withdraw huge volumes and return almost all (A consumes under 1%); recirculating towers withdraw little but evaporate most of it (B consumes 75%).
Plant A returns more than 99% of what it withdraws, the signature of once-through cooling.
Plant B consumes 75% of its withdrawal, which only an evaporative recirculating system does.
Dry (air) cooling consumes almost no water, but Plant B consumes 600 gal/MWh.
Answer every question to see your score
Pace it like exam day
The real exam gives you 5 h 20 min for 110 questions, a little under 3 minutes each. At exam speed the set fills close to 3 hours. Do it in two sittings, as the exam does: its break is 25 min, after about half the questions.
Keep the FE Reference Handbook open in a second window and look up every formula instead of recalling it. Search speed counts on exam day.
This practice exam and the NCEES one
This set is independent of NCEES, which sells its own practice exams (2 × 50 questions, $50 each), built on the same July 2020 specification. The real exam also uses alternative item types (multiple correct, point and click, drag and drop, fill in the blank); this set is multiple choice.
The NCEES practice exams are interactive products in its online store, not PDFs. Problem sets that circulate as PDFs are often older; check that their topics match the 15 areas of the specifications in force since July 2020.
How long is the real FE Environmental exam?
5 h 20 min of testing for 110 questions, a few of them unscored pretest items, inside an appointment of 6 h.
How close is this to the real exam?
The topics and their proportions follow the current specification; most problems here are calculations. The exam's own items are confidential, so no independent set, this one included, reproduces them.
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