PE Civil Water Resources and Environmental Exam Prep

Answer: B. 2.89 in.

Find the potential maximum retention first: S = 1000/CN − 10 = 1000/80 − 10 = 2.5 in. The initial abstraction is Ia = 0.2S = 0.5 in. Then Q = (P − Ia)² / (P − Ia + S) = (4.5)² / (4.5 + 2.5) = 20.25/7.0 = 2.89 in.

Why not the others: A sets Ia equal to S instead of 0.2S. C drops Ia entirely (P²/(P + S)). D is P − Ia, as if every drop after the initial abstraction ran off.

Try a change: Raise CN to 90 with the same storm. S drops to 1.11 in, Ia to 0.22 in, and Q rises to about 3.88 in. Higher CN means less retention and more runoff.

Source: USDA NRCS, TR-55 Urban Hydrology for Small Watersheds, 2nd ed. (June 1986), copy hosted by San Diego State University, Eqs. 2-1, 2-2 and 2-4, p. 2-1

Answer: B. 26%.

The annual exceedance probability is 1/100 = 0.01, so the chance of no exceedance in a year is 0.99. Over 30 independent years: risk = 1 − (0.99)³⁰ = 1 − 0.740 = 0.260, or about 26%.

Why not the others: A is the annual probability. C adds 1% per year for 30 years, which overstates the risk. D is the probability of no exceedance at all.

Source: Ventura County Public Works, Flood statistics: commonly asked questions (risk equation), Risk = 1 − (1 − 1/T)ⁿ, with a 50-year worked example

Answer: D. 13.5 min.

nL = 0.24 × 100 = 24. (24)^0.8 = 12.71. P₂^0.5 = 1.897 and s^0.4 = (0.02)^0.4 = 0.2091. Tt = 0.007 × 12.71 / (1.897 × 0.2091) = 0.224 hr = 13.5 min. The 100-ft length is within TR-55's sheet-flow limit of less than 300 ft.

Why not the others: A reports the 0.224 result as minutes instead of hours. B enters the slope as 2 (percent) instead of 0.02 ft/ft. C forgets the square root on P₂.

Source: USDA NRCS, TR-55 Urban Hydrology for Small Watersheds, 2nd ed. (June 1986), copy hosted by San Diego State University, Eq. 3-3 and Table 3-1 (dense grasses n = 0.24), p. 3-3; 300-ft limit, pp. 3-3 to 3-4

Answer: B. 0.350 m³/s; exceeds the drain capacity.

Weight C by area: C = (1.50 × 0.90 + 2.50 × 0.30)/4.00 = (1.35 + 0.75)/4.00 = 0.525. Then Q = 0.525 × 60 × 4.00/360 = 0.350 m³/s, which is 0.030 m³/s more than the drain can carry.

Why not the others: A slips a factor of 10 in the unit conversion. C averages the two coefficients (0.60) without weighting by area. D leaves out C entirely, as if all rain ran off.

Source: TxDOT Hydraulic Design Manual, Chapter 4, Section 12: Rational Method, Eq. 4-20 (Z = 360 for metric), Eq. 4-23 (mixed land use), and Assumptions and Limitations

Answer: D. Choose B: present worth about $95,100 vs. about $108,800 for A.

The uniform-series present worth factor is (P/A, 6%, 20) = [(1.06)²⁰ − 1] / [0.06(1.06)²⁰] = 11.4699. PW of A = 40,000 + 6,000 × 11.4699 = $108,820. PW of B = 55,000 + 3,500 × 11.4699 = $95,145. B is about $13,700 cheaper in present-worth terms.

Why not the others: A isn't supported by either calculation; the gap is about 13% of A's present worth. B looks only at first cost and ignores 20 years of O&M. C reaches the right choice but adds undiscounted dollars, which ignores the time value of money.

Source: NIST IR 85-3273-39, Energy Price Indices and Discount Factors for Life-Cycle Cost Analysis – 2024 (March 2024), Section 2.1, uniform present value factor, printed p. 5 (PDF p. 13); the 6% rate and project costs are supplied exercise inputs

Answer: C. 13 working days.

A finishes at day 2. B finishes at 2 + 6 = 8 and C at 2 + 4 = 6. D cannot start until both B and C finish, so it starts at day 8 and finishes at 11. E finishes at 13. The critical path, the longest path through the network, is A–B–D–E.

Why not the others: A stops when B finishes. B follows the shorter A–C–D–E path and forgets that D must wait for B. D adds every duration, as if B and C ran one after the other.

Source: GAO-16-89G, Schedule Assessment Guide (Dec. 2015), Critical path and total float definitions, pp. 4 and 6

Answer: B. 90.3 mm.

For a normally consolidated layer: Sc = [Cc·H₀/(1 + e₀)] · log₁₀(σ′f/σ′₀) = (0.18 × 3.0/1.80) × log₁₀(2) = 0.300 × 0.3010 = 0.0903 m = 90.3 mm.

Why not the others: A leaves out the layer thickness. C forgets to divide by 1 + e₀. D uses the natural log, but Cc is defined with log₁₀. None of these adds immediate or secondary settlement, which the question excludes.

Source: FHWA-NHI-06-088, Soils and Foundations Reference Manual, Vol. I (Dec. 2006), Section 7.5.2.1, Normally Consolidated Soils, p. 7-24

Answer: A. It does not meet the minimum; 1.5 is required.

EM 1110-2-1902, Table 3-1 lists a minimum of 1.5 for the downstream slope under long-term steady seepage at maximum storage pool (spillway crest or top of gates) for new earth and rock-fill dams. 1.42 is below that, so the slope needs redesign (for example, a flatter slope or better drainage) before it meets the criterion.

Why not the others: B uses the 1.3 minimum that Table 3-1 gives for end-of-construction conditions, not long-term steady seepage. C confuses the theoretical stability threshold (FS = 1.0) with a design minimum. D isn't a value the table gives for this condition.

Try a change: Same dam, end-of-construction condition, FS = 1.42. Now the slope exceeds the table's listed 1.3 minimum. However, footnote 2 says a higher minimum may be appropriate for a dam over 50 ft high on a soft foundation, or for a dam subjected to pool loading during construction. The loading condition and footnotes decide which criterion applies.

Source: USACE EM 1110-2-1902, Slope Stability (31 Oct. 2003), Table 3-1, Minimum Required Factors of Safety: New Earth and Rock-Fill Dams, printed p. 3-2, including footnote 2; Original USACE Table 3-1 reproduced in a TCU report, hosted by Brazil’s Câmara dos Deputados, Figure 15, printed/PDF p. 47, complete table and footnotes

Answer: A. 1,800 kg/m³; does not meet the requirement.

Dry density uses the dry mass: 2.70/0.00150 = 1,800 kg/m³. Relative compaction = 1,800/2,000 = 90%, which is below the 95% requirement.

Why not the others: B gets the density right but misreads the comparison. C and D use the moist mass (3.24 kg), which gives a wet density of 2,160 kg/m³, not dry density.

Source: FHWA-NHI-06-088, Soils and Foundations Reference Manual, Vol. I (Dec. 2006), Section 2.1.3 (weight-volume ratios), p. 2-5; Table 2-2, p. 2-8; Section 5.8.3 and Eq. 5-19 (relative compaction), pp. 5-76–5-77

Answer: 120 pcf.

For a saturated soil, S = 1, so Se = wGs gives e = 0.15 × 2.70 = 0.405. Dry unit weight γd = Gs·γw/(1 + e) = 2.70 × 62.4/1.405 = 119.9 pcf, or 120 pcf.

Common mistakes: Setting e equal to the water content (0.15) gives about 147 pcf. Computing the saturated unit weight, (Gs + e)γw/(1 + e), gives about 138 pcf, which answers a different question.

Source: FHWA-NHI-06-088, Soils and Foundations Reference Manual, Vol. I (Dec. 2006), Sections 2.1.3–2.1.4 and Table 2-2, Weight-volume relations, pp. 2-5 to 2-8

Answer: A. 1.0 × 10⁻² cm/s.

Darcy's law, Q = kA(h/L), rearranges to k = VL/(Aht), where V is the volume collected in time t. k = (500 × 20)/(80 × 40 × 300) = 10,000/960,000 = 0.0104 cm/s.

Why not the others: B flips the gradient (uses L/h instead of h/L). C leaves out the sample length. D forgets to divide by the 300-s test time.

Source: University of Connecticut, Groundwater Education and Research Program: Darcy Column Test for Permeability, Darcy-law equation and variable definitions below the column-test diagram, Q = K A dh/L; constant-head procedure, Q = Volume/time; FHWA-NHI-06-088, Soils and Foundations Reference Manual, Vol. I (Dec. 2006), Section 5.6, Permeability, p. 5-58

Answer: B. 24.0 m³/day.

150 mg/L = 0.150 kg/m³, so incoming solids = 8,000 × 0.150 = 1,200 kg/day. Captured solids = 0.60 × 1,200 = 720 kg/day. Each cubic meter of sludge holds 0.030 × 1,000 = 30.0 kg of dry solids, so sludge volume = 720/30.0 = 24.0 m³/day.

Why not the others: A applies the 60% capture twice. C sends all incoming solids to the sludge. D treats 3.0% as 0.003 instead of 0.030.

Source: MIT OpenCourseWare 1.061 Transport Processes in the Environment (Fall 2008), Lecture 2: Conservation of Mass, Integral conservation-of-mass statement, Eq. (4), reduced here to a steady balance

Answer: C. 707 gpd/ft².

Surface overflow (surface loading) rate = flow ÷ surface area. Each clarifier has area π(30)² = 2,827 ft², so the two total 5,655 ft². Rate = 4,000,000 gpd/5,655 ft² = 707 gpd/ft². Compare the result with the design criteria in the wastewater standard supplied for your exam rather than a remembered number.

Why not the others: A converts 4.0 MGD to 4,000 gpd instead of 4,000,000 gpd. B uses the 60-ft diameter as a radius. D sends all 4.0 MGD through one clarifier.

Source: Sacramento State College for EPA, Operation of Wastewater Treatment Plants: A Field Study Training Program, Surface Settling Rate or Surface Loading Rate, printed p. 5-34; flow in gpd divided by liquid surface area in ft²

Answer: C. 8.65 cfs.

Q = 3.33 × 4.0 × (0.75)^1.5 = 3.33 × 4.0 × 0.6495 = 8.65 cfs.

Why not the others: A raises H to the second power. B shortens the crest for two end contractions (L − 0.2H), which applies only to contracted weirs; this weir is suppressed. D raises H to the first power.

Source: USBR Water Measurement Manual, Chapter 7, Section 10: Standard Suppressed Rectangular Weir, Francis equation for the standard suppressed rectangular weir, Eq. 7-5

Answer: D. 17.4 kW.

The pump must add H = 12.0 + 8.0 = 20.0 m. Hydraulic (fluid) power = ρgQH = 1,000 × 9.81 × 0.060 × 20.0 = 11,772 W. Shaft power = 11,772/0.75 = 15,696 W. Electrical input = 15,696/0.90 = 17,440 W = 17.4 kW.

Why not the others: A multiplies by both efficiencies instead of dividing. B stops at hydraulic power. C stops at shaft power and ignores motor losses.

Source: U.S. DOE, Improving Pumping System Performance: A Sourcebook for Industry, 2nd ed. (May 2006), Fluid power and pump efficiency, printed p. 9 (PDF p. 13); shaft power divided by motor efficiency, Simple Calculation, printed p. 58 (PDF p. 62); FHWA HDS-4, Introduction to Highway Hydraulics, metric version (June 1997), Section 6.2, energy equation and losses

Answer: B. 10.2 ft.

d = 12 in = 1.0 ft, so d^4.871 = 1. h = 4.727 × 2,000 × 3.0^1.852 / 120^1.852 = 4.727 × 2,000 × 7.6494/7,089.958 ≈ 10.2 ft.

Why not the others: A uses 1,000 ft of pipe. C uses C = 100 (rougher pipe). D doubles the correct result.

Try a change: Double the flow to 6.0 cfs. Head loss rises by 2^1.852 = 3.6 times, to about 36.8 ft. Loss grows much faster than flow.

Source: U.S. EPA, EPANET 2.2 User Manual, Section 3.1, Table 3.1: Pipe Headloss Formulas for Full Flow, Table 3.1, Hazen-Williams resistance coefficient and flow exponent (feet, cfs)

Answer: C. 3.15 ft.

Area = π(1.0)²/4 = 0.7854 ft², so V = 2.5/0.7854 = 3.18 ft/s. h = f(L/D)V²/(2g) = 0.020 × (1,000/1.0) × (3.18)²/(2 × 32.2) = 0.020 × 1,000 × 0.157 = 3.15 ft. The same answer follows from EPANET's flow form, h = 0.0252 f L Q²/d⁵.

Why not the others: A is only the velocity head, V²/2g. B plugs the flow (2.5) in where the velocity belongs. D drops the 2 in 2g.

Source: U.S. EPA, EPANET 2.2 User Manual, Section 3.1, Table 3.1: Pipe Headloss Formulas for Full Flow, Table 3.1, Darcy-Weisbach resistance coefficient (feet, cfs); FHWA HDS-4, Introduction to Highway Hydraulics, metric version (June 1997), Section 6.3.1, friction losses

Answer: A. NPSHa ≈ 19.1 ft; cavitation risk.

For a suction lift, NPSHa = atmospheric head − static lift − friction − vapor pressure head = 33.9 − 12.0 − 2.0 − 0.84 = 19.06 ft. That is less than the 22 ft the pump requires, so it is at risk of cavitation, which occurs when pressure in the pump drops below the liquid's vapor pressure.

Why not the others: B adds the 12-ft lift as if the pump had a flooded suction. C ignores suction friction. D ignores vapor pressure. C and D still flag cavitation, but with the wrong NPSHa.

Source: Goulds Water Technology/Xylem, Water Products: Technical Data and Pump Fundamentals (Technical Manual TTECHWP R6), NPSHA for an open system with suction lift, Figure 4a, PDF p. 13; suction-pressure/velocity-head formulation, PDF p. 12; U.S. DOE, Improving Pumping System Performance: A Sourcebook for Industry, 2nd ed. (May 2006), Cavitation, printed p. 14; NPSH overview, printed p. 21; available-versus-required NPSH, printed pp. 38–39

Answer: C. 4.83 m³/s; below the required discharge.

A = 4.00 × 1.00 = 4.00 m². The wetted perimeter counts the bed and both walls: P = 4.00 + 2(1.00) = 6.00 m. R = A/P = 0.667 m. Q = (1/0.020)(4.00)(0.667)^(2/3)(0.0010)^(1/2) = 4.83 m³/s, about 0.17 m³/s short at this depth.

Why not the others: A uses S instead of √S. B includes the free surface in the wetted perimeter. D uses the depth (1.00 m) in place of the hydraulic radius.

Source: FHWA HDS-4, Introduction to Highway Hydraulics, metric version (June 1997), Section 4.3.1, Manning's equation and hydraulic radius, Eqs. 13, 15 and 17

Answer: B. 2.69 ft.

At critical flow, Q²/g = A³/T. For a rectangular section this reduces to yc = (q²/g)^(1/3) = (25²/32.2)^(1/3) = (19.41)^(1/3) = 2.69 ft.

Why not the others: A uses q instead of q². C is the minimum specific energy, Ec = 1.5yc. D stops before taking the cube root.

Source: Design Manual Chapter 2 Stormwater, Section 2F-2 Open Channel Flow (2013 ed.), hosted by Iowa State University InTrans, Eq. 2F-2.07, critical flow condition, p. 5; FHWA HDS-4, Introduction to Highway Hydraulics, metric version (June 1997), Section 4.6.2, critical depth in rectangular channels, Eq. 33

Answer: A. Fr = 2.01; E = 3.63 ft; supercritical.

V = Q/A = 120/(8.0 × 1.2) = 12.5 ft/s. For a rectangular channel the hydraulic depth A/T equals y, so Fr = V/√(gy) = 12.5/√(32.2 × 1.2) = 2.01. Fr > 1, so the flow is supercritical. Specific energy E = y + V²/2g = 1.2 + 12.5²/64.4 = 3.63 ft.

Why not the others: B gets the numbers right but reverses the regime: Fr greater than 1 means supercritical. C reports Fr² as Fr. D drops the 2 in 2g.

Try a change: If this flow enters a mild downstream channel, check the downstream water depth and control conditions. A hydraulic jump may form if they force a transition to subcritical flow; the mild slope alone does not guarantee it. For exams starting April 2027, FHWA HEC-14 is on the supplied reference list for jumps and energy dissipators.

Source: Design Manual Chapter 2 Stormwater, Section 2F-2 Open Channel Flow (2013 ed.), hosted by Iowa State University InTrans, Eq. 2F-2.05 (specific energy), p. 3; Eq. 2F-2.08 and Fr > 1 supercritical, p. 6; hydraulic-jump conditions, pp. 8–9

Answer: A. q = 0.072 m/day; 400 days.

Gradient i = 1.8/300 = 0.0060. Darcy flux q = Ki = 12.0 × 0.0060 = 0.072 m/day. Water actually moves through the pores faster: v = q/nₑ = 0.072/0.24 = 0.300 m/day. Travel time = 120/0.300 = 400 days.

Why not the others: B uses the Darcy flux as the travel speed. C labels the pore velocity as the Darcy flux. D multiplies by porosity instead of dividing, then uses that as a speed.

Source: EPA, EPACMTP Parameters/Data Background Document (April 2003), Section 5.3.5, definition of seepage velocity and Eq. 5.8, printed p. 5-49 (PDF p. 154); v = Ki/nₑ and Darcy flux = nₑv

Answer: B. 142 gpm.

Transmissivity T = Kb = 40 × 50 = 2,000 ft²/day. The Thiem equation gives Q = 2πT(h₂ − h₁)/ln(r₂/r₁) = 2π × 2,000 × 5.0/ln(10) = 62,832/2.303 = 27,290 ft³/day. Convert: 27,290 × 7.48/1,440 = 142 gpm.

Why not the others: A stops at cubic feet per minute. C uses log₁₀ in place of the natural log. D labels the ft³/day result as gpm.

Source: EPA, Handbook: Ground Water and Wellhead Protection (September 1994), Eq. 3-1, printed pp. 51–52, for transmissivity; Section 4.5.2, Eq. 4-15, printed p. 80, for the Thiem equilibrium relation. The two-observation form is derived by subtracting the drawdowns at the two radii

Answer: 5.12 acre-ft.

Volume released = S × area × head decline = 0.0004 × (2.0 × 640 acres) × 10 ft = 0.0004 × 1,280 × 10 = 5.12 acre-ft (about 223,000 ft³).

Common mistakes: Skipping the square-mile-to-acre conversion gives 0.008. Storativity in a confined aquifer is tiny, so a large head drop releases a modest volume.

Source: MIT OpenCourseWare 1.72 Groundwater Hydrology (Fall 2005), Lecture 7: Groundwater Storage, Specific-storage definition, p. 1, and S = Sₛb, p. 7; together give released volume = S × area × head decline

Answer: D. 78.0 mg/L.

At the limit, 1.20(18.0) + 0.300Ct = (1.20 + 0.300)(30.0). So 21.6 + 0.300Ct = 45.0 and Ct = 23.4/0.300 = 78.0 mg/L. Check: (21.6 + 23.4)/1.50 = 30.0 mg/L.

Why not the others: A applies the downstream target to the tributary. B weights the two flows equally. C finds the allowable increase above 18.0 mg/L but leaves off the baseline.

Source: MIT OpenCourseWare 1.061 Transport Processes in the Environment (Fall 2008), Lecture 2: Conservation of Mass, Integral conservation-of-mass statement, Eq. (4), reduced to a steady mixing balance

Answer: C. 263 mg/L.

BOD exerted by day t is BODt = L₀(1 − e^(−kt)). Solve for L₀: L₀ = 180/(1 − e^(−0.23 × 5)) = 180/(1 − 0.3166) = 180/0.6834 = 263 mg/L.

Why not the others: A assumes the BOD is finished after 5 days. B uses 10^(−kt), mixing a base-10 form with a base-e constant. D divides by e^(−kt) instead of by (1 − e^(−kt)).

Source: EPA, Technical Guidance Manual for Developing Total Maximum Daily Loads, Book II: Streams and Rivers, Part 1, Section 2.3.4.1, Eq. 2-5, printed p. 2-8 (PDF p. 25): exerted BOD = ultimate BOD × (1 − e^(−kt))

Answer: 3.7 mg/L.

Set dD/dt = 0 to find the critical time: tc = [1/(kr − kd)] ln{(kr/kd)[1 − D₀(kr − kd)/(kd·L₀)]} = 5 × ln(2 × 0.95) = 5 × ln(1.9) = 3.21 days. At that point the critical deficit is Dc = (kd/kr)·L₀·e^(−kd·tc) = 0.5 × 20 × e^(−0.642) = 5.26 mg/L. Minimum DO = 9.0 − 5.26 = 3.74, or 3.7 mg/L.

Common mistakes: One slip is entering the deficit (5.3) instead of the DO. Another is plugging tc into the deficit equation with a sign error; check that DO at tc is lower than DO at, say, 2 or 5 days.

Source: EPA, Technical Guidance Manual for Developing Total Maximum Daily Loads, Book II: Streams and Rivers, Part 1, Section 2.4.2, Eq. 2-27, printed p. 2-22 (PDF p. 39), and minimum-DO explanation, p. 2-24 (PDF p. 41); the critical-time formula is derived from this model. EPA uses Kₐ for the reaeration rate labeled kr here

Answer: A. 235 lb/day.

TMDL = wasteload allocations + load allocations + MOS. MOS = 0.10 × 400 = 40 lb/day. Wasteload allocation (WLA) = 5.0 MGD × 3.0 mg/L × 8.34 = 125.1 lb/day. Load allocation = 400 − 40 − 125.1 = 234.9, or about 235 lb/day.

Why not the others: B skips the margin of safety. C forgets the 8.34 conversion, so the plant's load comes out as only 15 lb/day. D subtracts the MOS but forgets the point source.

Source: 40 CFR 130.2, Definitions (eCFR), §130.2(g)–(i), LA, WLA and TMDL definitions; 40 CFR 130.7, TMDLs and individual water quality-based effluent limitations (eCFR), §130.7(c)(1), margin of safety; EPA, Preventive Maintenance for Small Public Water Systems Using Ground Water: Tasks and Logs, Commonly Used Formulas, printed/PDF p. 5; pounds per day = MGD × mg/L × 8.34

Answer: B. 50,000 gal per tank.

With one tank down, each operating tank gets 4,800,000/2 = 2,400,000 gal/day. Detention time = volume/flow, so V = 2,400,000 × (30/1,440) = 50,000 gal.

Why not the others: A spreads the flow over all three tanks and ignores the outage. C is the combined volume of the two operating tanks. D is the total installed volume of all three.

Source: EPA, Preventive Maintenance for Small Public Water Systems Using Ground Water: Tasks and Logs, Commonly Used Formulas: detention time = volume ÷ flow, p. 5

Answer: A. 51.8 mg·min/L.

Theoretical detention time = V/Q = 200,000 gal/(2,000,000/1,440 gpm) = 144 min. Contact time T (the T₁₀ time) = TDT × baffling factor = 144 × 0.30 = 43.2 min. CT = C × T = 1.2 × 43.2 = 51.8 mg·min/L.

Why not the others: B uses an average baffling factor (0.5) instead of the measured 0.30. C uses the full theoretical detention time and ignores short-circuiting. D multiplies the residual by the volume in thousands of gallons. A common slip not shown here is stopping at T = 43.2 min, which is a time, not a CT.

Source: EPA 815-R-20-003, Disinfection Profiling and Benchmarking Technical Guidance Manual (June 2020), Eq. 4-1 (CT), printed p. 24; Eq. 4-2 (TDT at peak hourly flow), p. 26; Table 4-2 (baffling factors), p. 27; Eq. 4-3 (T = TDT × BF), p. 28 (PDF pp. 34, 36–38)

Answer: D. 299 mg/L as CaCO₃.

Convert each ion to its CaCO₃ equivalent: hardness = 2.497(Ca) + 4.118(Mg) = 2.497 × 80 + 4.118 × 24 = 199.8 + 98.8 = 298.6, or about 299 mg/L as CaCO₃. (The factors are 50.04/20.04 for calcium and 50.04/12.15 for magnesium, ratios of equivalent weights.)

Why not the others: A adds the ion concentrations without converting. B counts calcium only. C uses magnesium's atomic weight (24.3) instead of its equivalent weight (12.15).

Source: EPA, 2021 MSGP Appendix J: Calculating Hardness, Section iii, mg/L CaCO₃ = 2.497(Ca) + 4.118(Mg), printed p. J-2 (PDF p. 2); cited for the chemistry equation

Answer: D. 8.0 MGD.

Future population = 25,000 × (1.02)²⁰ = 37,150. Average-day demand = 37,150 × 120 = 4.46 MGD. Maximum-day demand = 4.46 × 1.8 = 8.0 MGD.

Why not the others: A stops at average-day demand. B applies the factor to today's population. C grows the population linearly (25,000 × 1.40 = 35,000), which understates compound growth. Peaking factors vary by system; on the exam, use the one the problem gives you.

Source: Worked from the stated inputs (compound growth and the given peaking factor)

Answer: A and C.

A describes biological treatment: microorganisms mixed with the wastewater break down organic matter. C describes the return and waste of biomass after the secondary clarifier. You must select both A and C, and only those, to get credit, the same all-or-nothing rule NCEES uses.

Why not the others: B confuses biomass recycle with aeration; oxygen comes from mixing air into the aeration tank. D confuses settling suspended solids with removing dissolved material, which gravity can't do.

Source: EPA 832-R-04-001, Primer for Municipal Wastewater Treatment Systems (Sept. 2004), Secondary treatment and activated sludge, pp. 12–14; NCEES Examinee Guide, May 2026, Item formats and no partial credit, printed/PDF p. 11

Answer: B. 343 gpm.

Flowing full, R = A/P = D/4 = (8/12)/4 = 0.167 ft. V = (1.486/0.013)(0.167)^(2/3)(0.004)^(1/2) = 2.19 ft/s. A = π(0.667)²/4 = 0.349 ft², so Q = 0.764 cfs = 343 gpm. The 2.19 ft/s full-flow velocity also clears the 2.0 ft/s minimum in the 2014 Recommended Standards for Wastewater Facilities, which bases its minimum slopes on n = 0.013.

Why not the others: A leaves out the 1.486 constant. C uses R = D/2. D uses R = D. For a full circular pipe, R = πD²/4 ÷ πD = D/4.

Source: Design Manual Chapter 2 Stormwater, Section 2F-2 Open Channel Flow (2013 ed.), hosted by Iowa State University InTrans, Eq. 2F-2.01, Manning's equation in U.S. units, p. 2; Recommended Standards for Wastewater Facilities, 2014 ed. (GLUMRB), full-document mirror, Section 33.41, printed p. 30-1 (PDF p. 27): 2.0 ft/s full-flow minimum with n = 0.013; 8-in slope table, p. 30-2 (PDF p. 28)

Answer: B. 9.7 days.

SRT = solids in the system ÷ solids leaving per day. Inventory = 1.0 × 3,000 × 8.34 = 25,020 lb. Wasted = 0.025 × 10,000 × 8.34 = 2,085 lb/day; effluent = 4.0 × 15 × 8.34 = 500 lb/day. SRT = 25,020/2,585 = 9.7 days. (The 8.34 factors cancel: 3,000/(250 + 60).)

Why not the others: A is the hydraulic detention time, V/Q. C ignores the solids lost in the effluent. D ignores the wasted sludge.

Source: Operation of Wastewater Treatment Plants, 2nd ed., Vol. II (1980), California State University, Sacramento, hosted by EPA, Sections 11.73–11.74, MLSS aeration-basin inventory divided by wasted plus effluent solids, printed p. 63

Answer: C. 214 mg/L as CaCO₃.

Nitrification consumes about 7.14 mg of alkalinity (as CaCO₃) per mg of ammonia nitrogen oxidized: 7.14 × 30 = 214 mg/L. If influent alkalinity is lower than that plus a working residual, the plant may need to add alkalinity to keep pH stable.

Why not the others: A treats the ratio as 1:1. B uses 3.57, the alkalinity recovered by denitrification, not consumed by nitrification. D doubles the ratio.

Source: Long Island Sound Nitrogen Removal Training Program, Module 4 (SUNY Farmingdale, 1997), hosted by NYSDEC, Unit 1, transparencies 1-2 and 1-17 (7.14 lb as CaCO₃ per lb N oxidized); 1-19 (3.57 recovered)

Answer: C. 1,440 yd³ loose.

Bank volume = [(280 + 440)/2] × 90 = 32,400 ft³ = 32,400/27 = 1,200 yd³. Loose volume = 1,200 × 1.20 = 1,440 yd³.

Why not the others: A divides by 1.20 instead of multiplying. B stops at bank volume. D is the loose volume in cubic feet, mislabeled as cubic yards.

Source: INDOT, Earthworks, Chapter 6: Measurements and Earthwork Calculations, Volumes, average-end-area equation with the 2 × 27 divisor, printed p. 6-3 (PDF p. 4)

Answer: D. 34 ft.

Table B-1 allows 1½H:1V for Type C soil in excavations less than 20 ft deep. Each side runs 1.5 × 10 = 15 ft horizontally, so the top width is 4 + 2(15) = 34 ft.

Why not the others: A uses Type A soil (¾H:1V on both sides), or Type C on one side only. B uses Type B (1H:1V). C leaves out the 4-ft bottom. Sloping deeper than 20 ft must be designed by a registered professional engineer.

Source: OSHA, 29 CFR 1926 Subpart P, Appendix B: Sloping and Benching, Table B-1, Maximum Allowable Slopes, note 3; paragraphs (c)(3)(ii)–(iii), distress and surcharge conditions

Answer: A. 200 ft.

The low point is where the grade along the curve is zero: x = g₁L/(g₁ − g₂) = (−2.0)(500)/(−2.0 − 3.0) = 200 ft from the beginning of the curve.

Why not the others: B assumes the low point is at mid-curve, which only happens when the grades are equal and opposite. C gives the 300-ft distance measured back from the end of the curve. D doubles the correct value.

Source: University of Washington CIVE 316, Lecture 12: High or Low Points on a Curve, x = g₁L/(g₁ − g₂), distance from the BVC to the high or low point

Answer: D. 27,000 ft³.

Part 2.2.12.c of the 2022 CGP requires storage for either the calculated runoff from a 2-year, 24-hour storm or 3,600 ft³ per acre drained. Using the per-acre option: 3,600 × 7.5 = 27,000 ft³ (1,000 yd³).

Why not the others: A is the right quantity in cubic yards, mislabeled as cubic feet. B is the per-acre rate only. C halves the requirement. If your site is covered by a state construction stormwater permit instead of EPA's, check that permit's own basin criteria.

Source: EPA 2022 Construction General Permit (as modified), Part 2.2.12(c), sediment basin storage, printed p. 15 (PDF p. 17)