BetaThis feature is still being polished.Tell us what you think
NCEES Engineering Licensure

FE Environmental Practice Exam

FE Environmental exam practice — the NCEES discipline exam, your first step toward the PE.

Start Practising — Free

Built for FE Environmental candidates

  • NCEES FE Environmental topic mix: water & wastewater, air quality, solid & hazardous waste, environmental chemistry, and fate & transport
  • Answerable from the NCEES FE Reference Handbook + first principles
  • Mix of calculation and conceptual items, in SI and US customary units
  • Step-by-step solutions on every question (first 2 free; all with Premium)
  • Follow-up study chat to dig into any concept that trips you up

The FE Environmental exam is one of the seven NCEES Fundamentals of Engineering discipline exams — a 110-question, computer-based test, taken near graduation, that is the first step toward becoming a licensed Professional Engineer (PE). Every question is meant to be answered from the FE Reference Handbook and engineering fundamentals, so practice that drills the right Environmental topic mix, in the right units, with worked solutions is exactly how you build exam-day speed.

Crameleon builds your practice exam on demand — pick the pathway and module you're studying, and we generate a fresh paper with step-by-step solutions (the first 2 free; all of them with Premium). Every question comes with a follow-up study chat, so you can dig into any concept that trips you up without leaving the page. The generation takes under 2 minutes and your first exam is free.

Sample FE Environmental exam questions

Three original questions on FE Environmental exam topics, each with a worked solution.

Question 1Multiple choiceWater and Wastewater

A wastewater sample has a 5-day biochemical oxygen demand (BOD₅) of 180 mg/L at 20°C. The BOD reaction rate constant for the sample is 0.23 d⁻¹ (base e). Assuming first-order BOD kinetics, what is the ultimate carbonaceous BOD, L0L_0, of the sample?

  1. A.

    123 mg/L

  2. B.

    194 mg/L

  3. C.

    263 mg/L

  4. D.

    568 mg/L

Worked solution
Answer:C

With first-order kinetics, the BOD exerted after tt days is the fraction of the ultimate demand that has been oxidised by then:

BODt=L0(1−e−kt)\text{BOD}_t = L_0\left(1 - e^{-kt}\right)

The base-e form is used because the rate constant is stated as a base-e value.

Evaluate the fraction exerted at 5 days:

1−e−kt=1−e−(0.23)(5)=1−e−1.15=1−0.317=0.6831 - e^{-kt} = 1 - e^{-(0.23)(5)} = 1 - e^{-1.15} = 1 - 0.317 = 0.683

So 68.3% of the ultimate demand is exerted in the first 5 days. Solve for L0L_0:

L0=BOD51−e−kt=1800.683=263 mg/LL_0 = \frac{\text{BOD}_5}{1 - e^{-kt}} = \frac{180}{0.683} = 263\,\text{mg/L}

The ultimate BOD is larger than the 5-day BOD, as it must be, since 83 mg/L of demand remains to be exerted after day 5.

L₀ ≈ 263 mg/L (option C)

The most tempting wrong answer is B (194 mg/L), which uses 1−10−kt1-10^{-kt} with the base-e constant. The base-10 form is valid only with a base-10 constant, which would be 0.23/2.303=0.100.23/2.303 = 0.10 d⁻¹ and would give the same 263 mg/L. Option A (123 mg/L) multiplies by (1−e−kt)(1 - e^{-kt}) instead of dividing, and produces an ultimate BOD smaller than the 5-day BOD, which is impossible. Option D (568 mg/L) divides by e−kte^{-kt}, the fraction remaining, instead of the fraction exerted.

Question 2Short answerAir Quality and Control

An ambient air sample contains sulfur dioxide (SO₂, molar mass 64.06 g/mol) at a concentration of 0.075 ppm by volume. The sample is at 25°C and 1 atm, where the molar volume of an ideal gas is 24.45 L/mol. Express the SO₂ concentration in µg/m³.

Worked solution
Answer:

≈ 197 µg/m³

A volume concentration in ppm is a mole ratio for an ideal gas: 0.075 ppm means 0.075 mol of SO₂ per 10610^6 mol of air. Converting to a mass per volume requires the number of moles in a cubic metre of air at the stated temperature and pressure, which is why the molar volume is given.

Moles of air in 1 m³ (1,000 L):

nair=1,000 L24.45 L/mol=40.9 moln_{\text{air}} = \frac{1{,}000\,\text{L}}{24.45\,\text{L/mol}} = 40.9\,\text{mol}

Moles of SO₂ in that cubic metre:

nSO2=0.075×10−6×40.9=3.07×10−6 moln_{\text{SO}_2} = 0.075 \times 10^{-6} \times 40.9 = 3.07 \times 10^{-6}\,\text{mol}

Mass of SO₂, using the molar mass:

m=3.07×10−6 mol×64.06 g/mol=1.965×10−4 g=196.5 μgm = 3.07 \times 10^{-6}\,\text{mol} \times 64.06\,\text{g/mol} = 1.965 \times 10^{-4}\,\text{g} = 196.5\,\mu\text{g}

The steps collapse into the standard conversion used at 25°C and 1 atm:

C (μg/m3)=ppm×M×1,00024.45=0.075×64.06×1,00024.45=196.5C\,(\mu\text{g/m}^3) =\frac{\text{ppm} \times M \times 1{,}000}{24.45} = \frac{0.075 \times 64.06 \times 1{,}000}{24.45} = 196.5

SO₂ concentration ≈ 197 µg/m³

Question 3Worked problemGroundwater, Soils, and Sediments

A confined sand aquifer is 15 m thick and has a hydraulic conductivity of 2.5 m/d and an effective porosity of 0.25. Two observation wells are located 600 m apart along the direction of flow. The piezometric head is 32.0 m at the upgradient well and 30.5 m at the downgradient well. Assume one-dimensional, steady, uniform flow obeying Darcy's law.

(a) Determine the hydraulic gradient and the Darcy velocity (specific discharge).

(b) Determine the volumetric flow rate through a 400 m width of the aquifer, in m³/d.

(c) Determine the average linear (seepage) velocity of the groundwater and the time for a conservative, non-retarded tracer to travel from the upgradient well to the downgradient well, in years.

Worked solution
Answer:

(a) i = 0.0025, q = 6.25 × 10⁻³ m/d; (b) Q = 37.5 m³/d; (c) v = 0.025 m/d, t ≈ 24,000 d ≈ 66 years

(a) Gradient and Darcy velocity. The hydraulic gradient is the head loss per unit distance along the flow path:

i=ΔhL=32.0−30.5600=1.5600=0.0025i = \frac{\Delta h}{L} = \frac{32.0 - 30.5}{600} = \frac{1.5}{600} = 0.0025

Darcy's law gives the specific discharge, the flow per unit gross cross-sectional area of aquifer:

q=Ki=2.5(0.0025)=6.25×10−3 m/dq = Ki = 2.5(0.0025) = 6.25 \times 10^{-3}\,\text{m/d}

(b) Flow rate. Multiply the specific discharge by the gross area perpendicular to flow, which is the aquifer thickness times the width considered:

Q=qA=(6.25×10−3)(15×400)=37.5 m3/dQ = qA = (6.25 \times 10^{-3})(15 \times 400) = 37.5\,\text{m}^3\text{/d}

(c) Seepage velocity and travel time. Water moves only through the connected pore space, so the actual average velocity is the Darcy velocity divided by the effective porosity:

v=qne=6.25×10−30.25=0.025 m/dv = \frac{q}{n_e} = \frac{6.25 \times 10^{-3}}{0.25} = 0.025\,\text{m/d}

A conservative tracer moves with the water, so the travel time between the wells is the distance divided by the seepage velocity:

t=Lv=6000.025=24,000 d=24,000365=65.8 yearst = \frac{L}{v} = \frac{600}{0.025} = 24{,}000\,\text{d} = \frac{24{,}000}{365} = 65.8\,\text{years}

Using the Darcy velocity instead of the seepage velocity would overstate the travel time by a factor of 4, since the Darcy velocity is an area-averaged flux, not the speed of the water.

i = 0.0025, q = 6.25 × 10⁻³ m/d; Q = 37.5 m³/d; v = 0.025 m/d, travel time ≈ 24,000 d ≈ 66 years

Ready to practise?

Generate Your Free FE Environmental (Fundamentals of Engineering) Practice Exam

Your first exam is free — just sign in, no credit card.