The FE Environmental Engineering exam tests your knowledge across a wide spectrum of environmental topics — from water and wastewater treatment to groundwater hydrology, air quality modeling, risk assessment, and solid waste management. The best way to prepare is to work through realistic problems under exam-like conditions.

Below you will find 10 practice problems that mirror the style, difficulty, and topic distribution you will encounter on the actual NCEES FE Environmental exam. Each problem includes four answer choices and a detailed, step-by-step solution. Use these to identify your strengths, target your weaknesses, and build the problem-solving speed you need on test day.

Tip: Try solving each problem on your own before revealing the solution. Time yourself — you should average about 3 minutes per problem on the real exam.

Problem 1: Fundamental Principles — Mass Balance on a Mixing Junction

Two streams merge at a confluence. Stream A has a flow rate of 5 m³/s and a chloride concentration of 120 mg/L. Stream B has a flow rate of 3 m³/s and a chloride concentration of 40 mg/L. Assuming complete and instantaneous mixing with no reactions, what is the chloride concentration in the combined stream immediately downstream of the junction?

  • A) 80 mg/L
  • B) 90 mg/L
  • C) 95 mg/L
  • D) 100 mg/L
Show Solution

Answer: B

For a conservative substance (no reactions), we apply a steady-state mass balance at the junction:

\( Q_A \times C_A + Q_B \times C_B = Q_{mix} \times C_{mix} \)

Step 1: Calculate the total flow rate:

\( Q_{mix} = Q_A + Q_B = 5 + 3 = 8 \text{ m}^3/\text{s} \)

Step 2: Calculate the mass loading from each stream:

Mass from A = \( 5 \times 120 = 600 \text{ mg/L} \cdot \text{m}^3/\text{s} \)

Mass from B = \( 3 \times 40 = 120 \text{ mg/L} \cdot \text{m}^3/\text{s} \)

Step 3: Solve for the mixed concentration:

\( C_{mix} = \frac{600 + 120}{8} = \frac{720}{8} = \textbf{90 mg/L} \)

Note that the answer is not simply the average of 120 and 40 (which would be 80 mg/L). Because Stream A carries more flow, it has a greater influence on the mixed concentration. This flow-weighted average is the fundamental concept behind mass balance problems on the exam.

Problem 2: Water & Wastewater — BOD Kinetics

A wastewater sample has an ultimate BOD (L0) of 250 mg/L and a first-order BOD rate constant of k = 0.20 day−1 (base e). What is the 5-day BOD (BOD5) of this wastewater?

  • A) 125 mg/L
  • B) 158 mg/L
  • C) 182 mg/L
  • D) 210 mg/L
Show Solution

Answer: B

The BOD exerted at time t is given by:

\( L_t = L_0 (1 - e^{-kt}) \)

Step 1: Substitute the given values for t = 5 days:

\( L_5 = 250 (1 - e^{-0.20 \times 5}) \)

Step 2: Calculate the exponent:

\( -0.20 \times 5 = -1.0 \)

Step 3: Calculate the exponential term:

\( e^{-1.0} = 0.3679 \)

Step 4: Calculate BOD5:

\( L_5 = 250 (1 - 0.3679) = 250 \times 0.6321 = \textbf{158.0 mg/L} \)

The BOD5 is 158 mg/L, meaning that about 63% of the ultimate oxygen demand is exerted within 5 days. Note that answer choice A (125 mg/L) would be exactly half the ultimate BOD — a common trap for students who assume a linear relationship. Answer choice D (210 mg/L) represents about 84% of the ultimate BOD, which corresponds to a much longer time period. On the FE exam, always verify whether the rate constant k is given in base-e or base-10 — this changes which formula you use and is a classic source of errors.

Problem 3: Groundwater — Darcy’s Law

A confined aquifer is 12 m thick, 500 m wide, and has a hydraulic conductivity of K = 8.5 m/day. Two observation wells located 1,200 m apart show hydraulic heads of 42.0 m and 38.4 m, respectively. What is the volumetric flow rate through the aquifer cross-section?

  • A) 102 m³/day
  • B) 153 m³/day
  • C) 204 m³/day
  • D) 306 m³/day
Show Solution

Answer: B

Darcy’s law states: \( Q = KiA \), where K is hydraulic conductivity, i is the hydraulic gradient, and A is the cross-sectional area perpendicular to flow.

Step 1: Calculate the hydraulic gradient:

\( i = \frac{\Delta h}{L} = \frac{42.0 - 38.4}{1200} = \frac{3.6}{1200} = 0.003 \)

Step 2: Calculate the cross-sectional area of the aquifer:

\( A = \text{thickness} \times \text{width} = 12 \times 500 = 6{,}000 \text{ m}^2 \)

Step 3: Apply Darcy’s law:

\( Q = KiA = 8.5 \times 0.003 \times 6{,}000 \)

\( Q = 8.5 \times 18 = \textbf{153 m}^3\textbf{/day} \)

This is the Darcy flux (specific discharge) times the area. Remember that the actual seepage velocity of water through the pore spaces would be higher: \( v_s = Q / (A \times n_e) \), where \( n_e \) is the effective porosity. The exam frequently tests the distinction between Darcy velocity and seepage velocity.

Problem 4: Air Quality — Gaussian Plume Dispersion

A factory stack emits SO2 at a rate of Q = 50 g/s. The effective stack height is H = 40 m. At a downwind distance of 800 m, the horizontal and vertical dispersion coefficients are σy = 100 m and σz = 40 m, and the wind speed at stack height is u = 5 m/s. Using the Gaussian plume model for ground-level centerline concentration:

\( C = \dfrac{Q}{\pi\, u\, \sigma_y\, \sigma_z} \exp\!\left(-\dfrac{H^2}{2\,\sigma_z^2}\right) \)

what is the SO2 concentration at ground level on the plume centerline?

  • A) 0.20 mg/m³
  • B) 0.48 mg/m³
  • C) 0.80 mg/m³
  • D) 1.27 mg/m³
Show Solution

Answer: B

Apply the Gaussian plume formula for ground-level centerline concentration (y = 0, z = 0) with ground reflection:

\( C = \frac{Q}{\pi u \sigma_y \sigma_z} \exp\left(-\frac{H^2}{2\sigma_z^2}\right) \)

Step 1: Calculate the pre-exponential term:

\( \frac{Q}{\pi u \sigma_y \sigma_z} = \frac{50}{\pi \times 5 \times 100 \times 40} = \frac{50}{62{,}832} = 7.958 \times 10^{-4} \text{ g/m}^3 \)

Step 2: Calculate the exponential argument:

\( \frac{H^2}{2\sigma_z^2} = \frac{40^2}{2 \times 40^2} = \frac{1{,}600}{3{,}200} = 0.50 \)

Step 3: Evaluate the exponential:

\( \exp(-0.50) = 0.6065 \)

Step 4: Multiply to get the final concentration:

\( C = 7.958 \times 10^{-4} \times 0.6065 = 4.826 \times 10^{-4} \text{ g/m}^3 \)

Step 5: Convert to mg/m³:

\( C = \textbf{0.48 mg/m}^3 \)

The exponential term reduces the pre-exponential concentration by about 40% because the effective stack height H equals σz. If H were much larger than σz, the exponential term would approach zero and the ground-level concentration would be negligible. Answer choice C (0.80 mg/m³) is approximately the pre-exponential term alone — a common error when students forget the height-dependent exponential factor. On the FE exam, always check that your answer includes both the dispersion term and the height correction.

Problem 5: Health Hazards & Risk Assessment — Cancer Risk Calculation

A resident is exposed to a carcinogenic contaminant in drinking water at a concentration of 0.05 mg/L. The ingestion rate is 2 L/day, exposure frequency is 350 days/year, exposure duration is 30 years, body weight is 70 kg, and the averaging time for carcinogens is 70 years. The oral cancer slope factor is SF = 0.3 (mg/kg/day)−1. What is the excess lifetime cancer risk?

  • A) 1.8 × 10−4
  • B) 2.6 × 10−4
  • C) 5.5 × 10−5
  • D) 8.2 × 10−5
Show Solution

Answer: A

The cancer risk is calculated as Risk = CDI × SF, where CDI is the chronic daily intake.

Step 1: Calculate the chronic daily intake (CDI):

\( \text{CDI} = \frac{C \times IR \times EF \times ED}{BW \times AT} \)

Where: C = 0.05 mg/L, IR = 2 L/day, EF = 350 days/year, ED = 30 years, BW = 70 kg, AT = 70 years × 365 days/year = 25,550 days

\( \text{CDI} = \frac{0.05 \times 2 \times 350 \times 30}{70 \times 25{,}550} \)

Step 2: Calculate the numerator:

\( 0.05 \times 2 = 0.1 \)

\( 0.1 \times 350 = 35 \)

\( 35 \times 30 = 1{,}050 \text{ mg-days/L} \)

Step 3: Calculate the denominator:

\( 70 \times 25{,}550 = 1{,}788{,}500 \text{ kg-days} \)

Step 4: Calculate CDI:

\( \text{CDI} = \frac{1{,}050}{1{,}788{,}500} = 5.872 \times 10^{-4} \text{ mg/kg/day} \)

Step 5: Calculate the cancer risk:

\( \text{Risk} = \text{CDI} \times \text{SF} = 5.872 \times 10^{-4} \times 0.3 = \textbf{1.76} \times \textbf{10}^{-4} \)

This rounds to approximately \( 1.8 \times 10^{-4} \), which exceeds the commonly used regulatory threshold of \( 10^{-4} \) (1 in 10,000). This level of risk would typically trigger the need for remediation or an alternative water supply. On the exam, pay careful attention to the averaging time — for carcinogens it is always 70 years (lifetime), while for non-carcinogens it equals the exposure duration.

Problem 6: Water & Wastewater — Activated Sludge Design

An activated sludge system treats 10,000 m³/day of wastewater with an influent BOD5 of 200 mg/L. The aeration tank volume is 2,500 m³ and the mixed liquor volatile suspended solids (MLVSS) concentration is 3,000 mg/L. What is the food-to-microorganism (F/M) ratio?

  • A) 0.13 day−1
  • B) 0.27 day−1
  • C) 0.40 day−1
  • D) 0.67 day−1
Show Solution

Answer: B

The food-to-microorganism ratio is defined as:

\( F/M = \frac{Q \times S_0}{V \times X} \)

Where: Q = flow rate, S0 = influent substrate concentration, V = aeration tank volume, X = MLVSS concentration.

Step 1: Substitute the given values:

\( F/M = \frac{10{,}000 \text{ m}^3/\text{day} \times 200 \text{ mg/L}}{2{,}500 \text{ m}^3 \times 3{,}000 \text{ mg/L}} \)

Step 2: Calculate the numerator:

\( 10{,}000 \times 200 = 2{,}000{,}000 \)

Step 3: Calculate the denominator:

\( 2{,}500 \times 3{,}000 = 7{,}500{,}000 \)

Step 4: Calculate F/M:

\( F/M = \frac{2{,}000{,}000}{7{,}500{,}000} = \textbf{0.27 day}^{-1} \)

An F/M ratio of 0.27 day−1 falls within the typical range for a conventional activated sludge system (0.2–0.5 day−1). Lower F/M ratios correspond to extended aeration systems, while higher values indicate high-rate systems. The exam may also ask you to calculate the solids retention time (SRT), which is inversely related to F/M for a given system.

Problem 7: Surface Water & Hydrology — Manning’s Equation

A rectangular concrete-lined open channel is 3 m wide with a uniform flow depth of 1.5 m. The channel bed slope is S = 0.002 and the Manning’s roughness coefficient is n = 0.013. What is the discharge? Use Manning’s equation (SI): \( Q = \frac{1}{n} A R^{2/3} S^{1/2} \)

  • A) 9.8 m³/s
  • B) 12.8 m³/s
  • C) 15.5 m³/s
  • D) 18.2 m³/s
Show Solution

Answer: B

Step 1: Calculate the cross-sectional area of flow for a rectangular channel:

\( A = b \times y = 3 \times 1.5 = 4.5 \text{ m}^2 \)

Step 2: Calculate the wetted perimeter:

\( P = b + 2y = 3 + 2(1.5) = 3 + 3 = 6.0 \text{ m} \)

Step 3: Calculate the hydraulic radius:

\( R = A / P = 4.5 / 6.0 = 0.75 \text{ m} \)

Step 4: Calculate \( R^{2/3} \):

\( R^{2/3} = (0.75)^{2/3} = 0.8255 \)

Step 5: Calculate \( S^{1/2} \):

\( S^{1/2} = (0.002)^{1/2} = 0.04472 \)

Step 6: Apply Manning’s equation:

\( Q = \frac{1}{0.013} \times 4.5 \times 0.8255 \times 0.04472 \)

First: \( A \times R^{2/3} = 4.5 \times 0.8255 = 3.715 \)

Then: \( 3.715 \times 0.04472 = 0.1661 \)

Finally: \( Q = 0.1661 / 0.013 = \textbf{12.8 m}^3\textbf{/s} \)

Manning’s equation is one of the most commonly used formulas on the FE Environmental exam. For rectangular channels, the geometry calculations are straightforward, but watch for trapezoidal or circular cross-sections where the area and wetted perimeter formulas are more involved. The roughness coefficient n = 0.013 is typical for finished concrete; natural channels with vegetation or irregularities have much higher n values (0.03–0.05), which significantly reduce discharge capacity.

Problem 8: Solid & Hazardous Waste — Landfill Leachate Generation

A municipal solid waste landfill has an active area of 5 hectares (50,000 m²). The annual precipitation is 1,200 mm, of which 25% percolates through the waste to become leachate. Evapotranspiration and runoff account for the remainder. What is the annual leachate generation rate?

  • A) 7,500 m³/year
  • B) 15,000 m³/year
  • C) 30,000 m³/year
  • D) 60,000 m³/year
Show Solution

Answer: B

Leachate generation is calculated using a water balance approach:

\( L = P \times A \times \text{PERC} \)

Where P = precipitation, A = area, and PERC = fraction that percolates.

Step 1: Convert precipitation to meters:

\( P = 1{,}200 \text{ mm} = 1.2 \text{ m/year} \)

Step 2: Calculate the total volume of precipitation falling on the landfill:

\( V_{precip} = P \times A = 1.2 \times 50{,}000 = 60{,}000 \text{ m}^3/\text{year} \)

Step 3: Calculate the leachate volume (25% percolation):

\( L = 0.25 \times 60{,}000 = \textbf{15,000 m}^3\textbf{/year} \)

This is approximately 41 m³/day of leachate that must be collected and treated. In practice, the percolation fraction depends on cover design, waste density, and climate. The EPA HELP (Hydrologic Evaluation of Landfill Performance) model is used for more detailed analysis. On the FE exam, these problems are straightforward water balance calculations.

Problem 9: Water & Wastewater — CSTR Reactor Design

A completely mixed reactor (CSTR) is used as a disinfection contact chamber. The influent pathogen concentration is 10,000 organisms/L, the first-order inactivation rate constant is k = 0.5 min−1, and the required effluent concentration is 100 organisms/L. What is the minimum hydraulic detention time needed?

  • A) 9.2 min
  • B) 99 min
  • C) 198 min
  • D) 396 min
Show Solution

Answer: C

For a CSTR with first-order decay, the effluent concentration is:

\( C_{out} = \frac{C_{in}}{1 + k\tau} \)

Step 1: Rearrange to solve for the detention time \( \tau \):

\( 1 + k\tau = \frac{C_{in}}{C_{out}} \)

\( k\tau = \frac{C_{in}}{C_{out}} - 1 \)

\( \tau = \frac{1}{k}\left(\frac{C_{in}}{C_{out}} - 1\right) \)

Step 2: Substitute the given values:

\( \tau = \frac{1}{0.5}\left(\frac{10{,}000}{100} - 1\right) \)

\( \tau = 2 \times (100 - 1) = 2 \times 99 = \textbf{198 min} \)

Note how much detention time the CSTR requires for 2-log removal (99% removal). For comparison, a plug flow reactor (PFR) with the same k would achieve the same removal in: \( \tau = -\ln(C_{out}/C_{in})/k = -\ln(0.01)/0.5 = 4.605/0.5 = 9.2 \) min. This dramatic difference (198 min vs. 9.2 min) illustrates why PFRs are far more efficient than CSTRs for disinfection. Answer choice A corresponds to the PFR solution — a common trap on the exam.

Problem 10: Energy & Environment — Power Plant Efficiency and Emissions

A coal-fired power plant operates at a thermal efficiency of 35% and produces 500 MW of electrical power. The coal has a heating value of 25,000 kJ/kg and a carbon content of 65% by mass. Assuming complete combustion, what is the approximate rate of CO2 emissions? (Molecular weights: C = 12, O = 16, CO2 = 44)

  • A) 136 kg/s
  • B) 109 kg/s
  • C) 87 kg/s
  • D) 57 kg/s
Show Solution

Answer: A

Step 1: Calculate the total heat input required:

\( Q_{in} = \frac{P_{out}}{\eta} = \frac{500 \text{ MW}}{0.35} = 1{,}428.6 \text{ MW} = 1{,}428{,}600 \text{ kJ/s} \)

Step 2: Calculate the coal consumption rate:

\( \dot{m}_{coal} = \frac{Q_{in}}{HV} = \frac{1{,}428{,}600}{25{,}000} = 57.14 \text{ kg/s} \)

Step 3: Calculate the carbon combustion rate:

\( \dot{m}_C = 0.65 \times 57.14 = 37.14 \text{ kg/s} \)

Step 4: Convert carbon to CO2 using stoichiometry:

C + O2 → CO2

12 kg C produces 44 kg CO2

\( \dot{m}_{CO_2} = 37.14 \times \frac{44}{12} = 37.14 \times 3.667 = \textbf{136.1 kg/s} \)

This corresponds to approximately 4.3 million tonnes of CO2 per year if the plant runs continuously. This type of problem combines thermodynamics, stoichiometry, and environmental impact assessment — a classic cross-topic FE Environmental question. Note that answer choice D (57 kg/s) is the coal consumption rate, not the CO2 rate — another common trap.

Tips for Using These Practice Problems

Working through practice problems is one of the highest-value activities in your FE exam preparation, but how you practice matters just as much as how much you practice. Here are some tips to get the most out of these problems:

What Topics Does the FE Environmental Exam Cover?

The FE Environmental Engineering exam, administered by NCEES, covers 15 major topic areas:

The 10 problems above span many of these topic areas. To fully prepare, you need to practice across all of them — especially Water & Wastewater, Surface Water & Hydrology, and Groundwater, which together can account for roughly 25–37% of the exam.

Next Steps

These 10 problems give you a solid starting point, but the FE Environmental exam covers far more ground. To build true exam readiness, you need hundreds of problems across every topic, with instant feedback and detailed explanations for every answer choice.

Disclaimer: This content is for educational purposes only and is not affiliated with, endorsed by, or sponsored by NCEES. “FE” and “Fundamentals of Engineering” are trademarks of NCEES. Always refer to the official NCEES website for the most current exam specifications and policies.