Water resources and environmental is one of the broader knowledge areas on the FE Civil exam: 10 to 15 questions spanning open channel flow, pressure conduits, pump curves, hydrology, and water and wastewater treatment. The problems below are the kind you should be able to finish in under three minutes each, using only the NCEES FE Reference Handbook.
Each problem is followed by a full solution showing which handbook equation was used and where the unit conversions happen. Work them without looking at the answer first. If you can set up Manning's equation, a Hazen-Williams headloss, and a rational-method peak flow without hesitating, the exam version of these will not surprise you.
Exam weight: NCEES lists Water Resources and Environmental Engineering at 10-15 questions of the 110-question FE Civil exam. Work each problem below on paper first, then reveal the worked solution — reading a solution you have not attempted builds recognition, not recall.
What NCEES Tests in Water Resources and Environmental Engineering
The NCEES specification for Water Resources and Environmental Engineering covers hydrology basics such as rainfall-runoff, hydrographs and the rational method; hydraulics of closed conduits including energy and continuity, friction and minor losses, and pump power and system curves; and open channel flow through Manning's equation, normal and critical depth, specific energy, weirs and orifices. Expect 8 to 12 of the 110 questions from this area.
The environmental half draws on water quality, drinking water treatment and distribution, wastewater collection and treatment, and stormwater management. Typical items ask for a detention time, a chlorine dose after demand is satisfied, a BOD loading or removal efficiency, or the sizing of a sedimentation basin. Groundwater appears as well, usually as Darcy's law or a straightforward well drawdown calculation.
5 Free Water Resources and Environmental Engineering Practice Problems
Each water resources and environmental problem below comes from the FEExamClick FE Civil question bank, with worked solutions, and matches the style, difficulty and format of the real exam. Attempt each one under a three-minute limit — roughly the pace the exam demands.
Problem 1 — K. Water and wastewater treatment (e.g., biological processes, softening, drinking water treatment)
A facultative pond is \( 250\,\mathrm{m} \) long, \( 120\,\mathrm{m} \) wide, and operates at an average water depth of \( 1.5\,\mathrm{m} \). Average flow is \( 3{,}600\,\mathrm{m}^3/\mathrm{day} \). What is the pond detention time, most nearly?
Answer: C) \( 12.5\,\mathrm{days} \)
Step 1 – Identify given information:
$$Volume V = L·W·d = 250·120·1.5 = 45,000 m^{3}$$
Step 2 – Apply formula:
$$HRT = V/Q = 45000/3600 = 12.5 days$$
Correct answer: (C)
Problem 2 — B. Basic hydraulics (e.g., Manning equation, Bernoulli theorem, open-channel flow)
A sharp-edged orifice of diameter \( 50\,\mathrm{mm} \) discharges water from a large tank under head \( h = 2.0\,\mathrm{m} \). Assume \( C_d = 0.61 \). What is the discharge \( Q \)?
Answer: C) \( 0.0075\,\mathrm{m}^3/\mathrm{s} \)
Step 1 – Identify given information:
Orifice diameter \( d = 50 \) mm = 0.05 m, head \( h = 2.0 \) m, \( C_d = 0.61 \).
Step 2 – Orifice discharge formula (FE Handbook):
$$Q = C_d \cdot A \cdot \sqrt{2gh}$$
Step 3 – Solve:
$$A = \frac{\pi (0.05)^2}{4} = 0.0019635 \;\text{m}^2$$
$$\sqrt{2gh} = \sqrt{2(9.81)(2.0)} = \sqrt{39.24} = 6.264 \;\text{m/s}$$
$$Q = 0.61 \times 0.0019635 \times 6.264 = 0.0075 \;\text{m}^3\text{/s}$$
Step 4 – Answer: \( Q \approx 0.0075 \) m\(^3\)/s. The discharge coefficient \( C_d \approx 0.61{-}0.62 \) is typical for sharp-edged orifices (vena contracta). Option (D) \( 0.0123 \) omits \( C_d \); (B) \( 0.0053 \) uses \( \sqrt{gh} \) instead of \( \sqrt{2gh} \); (A) \( 0.0019 \) mistakenly uses the radius in \( \pi d^2/4 \).
Correct answer: (C)
Problem 3 — H. Groundwater (e.g., flow, wells, drawdown)
A pumping well in a confined aquifer has discharge \( Q = 0.012\,\mathrm{m}^3/\mathrm{s} \). The aquifer transmissivity is \( T = 2.5 \times 10^{-3}\,\mathrm{m}^2/\mathrm{s} \). What is the drawdown difference (\( s_1 - s_2 \)) between radii \( r_1 = 20\,\mathrm{m} \) and \( r_2 = 200\,\mathrm{m} \) under steady-state conditions (Thiem equation)?
Answer: A) \( 1.8\,\mathrm{m} \)
Step 1 – Identify given information:
\( Q = 0.012 \) m\(^3\)/s, \( T = 2.5 \times 10^{-3} \) m\(^2\)/s, \( r_1 = 20 \) m, \( r_2 = 200 \) m.
Step 2 – Thiem equation for confined aquifer (FE Handbook):
$$s_1 - s_2 = \frac{Q}{2\pi T} \ln\left(\frac{r_2}{r_1}\right)$$
Step 3 – Solve:
\( s_1 - s_2 = \frac{0.012}{2\pi(2.5 \times 10^{-3})} \ln\left(\frac{200}{20}\right) = \frac{0.012}{0.01571} \times \ln(10) = 0.7639 \times 2.3026 = 1.759 \) m
Step 4 – Answer:
The drawdown difference is \( \approx 1.8\,\text{m} \). Because the equation is logarithmic, this number depends only on the RATIO \( r_2/r_1 \): every ten-fold step outward from the well costs the same \( Q\ln(10)/(2\pi T) \), whether it is from 2 m to 20 m or from 20 m to 200 m.
(D) \( 11\,\text{m} \) drops the \( 2\pi \) from the denominator entirely, computing \( Q\ln(r_2/r_1)/T \), which inflates the answer by a factor of \( 6.28 \). It is the one real trap on the list.
(B) \( 3.6\,\text{m} \) and (C) \( 5.4\,\text{m} \) stand in for dropping only half of that constant: using \( \pi T \) gives \( 3.5\,\text{m} \) and using \( 2T \) gives \( 5.5\,\text{m} \). Writing the constant out in full before substituting is what prevents all three.
Correct answer: (A)
Problem 4 — A. Basic hydrology (e.g., infiltration, rainfall, runoff, watersheds)
A watershed has \( CN = 92 \) and receives rainfall depth \( P = 2.0\,\mathrm{in} \). What is the direct runoff depth \( Q \) (SCS CN method)?
Answer: B) \( 1.2\,\mathrm{in} \)
Step 1 – Identify given information:
\( CN = 92 \), rainfall depth \( P = 2.0 \) in.
Step 2 – SCS Curve Number formulas (FE Handbook):
\( S = \frac{1000}{CN} - 10 \), \( I_a = 0.2S \), \( Q = \frac{(P - I_a)^2}{P - I_a + S} \)
Step 3 – Solve step by step:
$$S = \frac{1000}{92} - 10 = 0.870 \;\text{in}$$
$$I_a = 0.2 \times 0.870 = 0.174 \;\text{in}$$
Since \( P = 2.0 > I_a = 0.174 \):
$$Q = \frac{(2.0 - 0.174)^2}{(2.0 - 0.174) + 0.870} = \frac{(1.826)^2}{1.826 + 0.870} = \frac{3.334}{2.696} = 1.237 \;\text{in}$$
Step 4 – Answer: \( Q \approx 1.2 \) in. With CN = 92, about 62% of the 2-inch rainfall becomes direct runoff. The small initial abstraction (0.17 in) reflects highly impervious conditions.
Correct answer: (B)
Problem 5 — B. Basic hydraulics (e.g., Manning equation, Bernoulli theorem, open-channel flow)
A rectangular open channel has width \( b = 3.0 \) m and flow depth \( y = 1.2 \) m. Manning \( n = 0.013 \) and slope \( S = 0.002 \). Estimate the discharge \( Q \) using Manning's equation (SI).
Answer: D) \( 9.5\,\mathrm{m}^3/\mathrm{s} \)
Step 1 – Identify given information:
| Parameter | Value |
|---|---|
| Width \( b \) | 3.0 m |
| Depth \( y \) | 1.2 m |
| Manning \( n \) | 0.013 |
| Slope \( S \) | 0.002 |
Step 2 – Manning's equation (SI):
\( Q = \frac{1}{n} A R_h^{2/3} S^{1/2} \) where \( A = by \), \( P = b + 2y \), \( R_h = A/P \).
Step 3 – Solve:
$$A = 3.0 \times 1.2 = 3.6 \;\text{m}^2$$
$$P = 3.0 + 2(1.2) = 5.4 \;\text{m}$$
$$R_h = \frac{3.6}{5.4} = 0.667 \;\text{m}$$
$$Q = \frac{1}{0.013} \times 3.6 \times (0.667)^{2/3} \times (0.002)^{1/2} = 76.92 \times 3.6 \times 0.763 \times 0.04472 = 9.45 \;\text{m}^3\text{/s}$$
Step 4 – Answer: \( Q \approx 9.5 \) m\(^3\)/s. In SI units, Manning's equation has the factor \( 1/n \). In US customary units it becomes \( 1.486/n \).
Counting the open water surface as part of the wetted perimeter inflates P to 8.4 m, shrinking the hydraulic radius and the discharge with it.
Correct answer: (D)
Using the FE Reference Handbook for Water Resources and Environmental Engineering
The water section of the handbook is long, so navigate by search term, not by scrolling. Search Manning to land in open channel flow, Hazen for pipe friction, and Darcy carefully, since it returns both Darcy-Weisbach and Darcy's law for groundwater. Learn where the Moody diagram sits; the n and C tables are next to the equations that use them.
Four Mistakes That Cost Points
- Using the wrong Manning constant for the unit system Manning's equation carries a different constant in each system: 1.486 in US customary units, 1.0 in SI. Drop the 1.486 in a problem stated in feet and seconds and your velocity is low by a factor of about 1.5, which usually still lands on one of the four choices.
- Treating hydraulic radius as flow depth R equals area over wetted perimeter, not depth. For a wide rectangular channel R approaches y, which is why the shortcut feels safe, but for a partly full circular pipe or a trapezoidal section it is nowhere close. Sketch the section and compute A and P explicitly every time.
- Confusing chlorine dose, demand and residual Dose equals demand plus residual. Problems hand you two of the three and ask for the third, and candidates who read dose as residual get a plausible wrong number. Write the three-term relationship on your scratch board before you touch any of the given values.
- Rational method unit factors Q equals CiA gives cubic feet per second directly when i is in inches per hour and A in acres, because the conversion factor is about 1.008. In SI you need the 1/360 factor with i in mm/hr and A in hectares. Confirm which system the problem uses before converting anything.
Frequently Asked Questions
How many water resources questions are on the FE Civil exam?
NCEES specifies 10 to 15 questions for Water Resources and Environmental Engineering out of the 110 on the exam. It sits alongside structural and geotechnical as one of the larger knowledge areas, so it deserves real preparation time even if your coursework leaned toward another specialty.
Do I need to memorize Manning's n or Hazen-Williams C values?
No. Both tables are in the NCEES FE Reference Handbook, which is on screen and searchable during the exam. What you do need is to recognize which coefficient a given problem is asking for and to know where the table lives, because hunting for it costs a minute you do not have.
How much treatment process detail does the exam test?
Less than a treatment course covers. Questions stay at the level of loading rates, detention times, removal efficiencies and basic dosing, which is arithmetic on a defined flow or volume. You are not asked to design a process train or recall proprietary equipment. If you can run a mass balance on a reactor, most items are reachable.
What is the fastest way to improve on open channel problems?
Practice identifying the unknown before writing anything. Most open channel items are Manning's equation solved for one of Q, n, S or a geometric quantity, plus a specific energy or critical depth check. Once you can classify an item in ten seconds the arithmetic is short, and mixed-problem repetition builds that reflex faster than rereading theory.
Keep Going
Done with water resources and environmental? Work through the other FE Civil knowledge areas with more free practice problems, review the full FE Civil exam topic breakdown, or plan your preparation with our FE exam study guide and study timeline. Or test yourself against the clock with the free FE Civil practice exam: 20 timed questions in the CBT format.