Water resources and environmental is one of the broader knowledge areas on the FE Civil exam: 8 to 12 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 8-12 questions (7-11%) 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 problem below comes from the PECivilClick FE Civil question bank 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
Two adjacent drainage areas flow to the same inlet: Area 1 = \( 3\,\mathrm{ac} \) with \( C = 0.30 \) and Area 2 = \( 5\,\mathrm{ac} \) with \( C = 0.70 \). For a storm intensity of \( 4.0\,\mathrm{in/hr} \), what is the peak runoff \( Q \) using the rational method?
A) \( 9\,\mathrm{ft}^3/\mathrm{s} \)
B) \( 14\,\mathrm{ft}^3/\mathrm{s} \)
C) \( 18\,\mathrm{ft}^3/\mathrm{s} \)
D) \( 26\,\mathrm{ft}^3/\mathrm{s} \)
Answer: C) \( 18\,\mathrm{ft}^3/\mathrm{s} \)
Step 1 – Identify the given information:
| Area | Size (ac) | Runoff Coeff. (C) |
|---|---|---|
| 1 | 3 | 0.30 |
| 2 | 5 | 0.70 |
| Total | 8 | — |
Storm intensity: \( i = 4.0 \;\text{in/hr} \)
Step 2 – Recall the Rational Method formula (from the FE Handbook):
\( Q = C \cdot i \cdot A \)
where \( Q \) = peak runoff (ft\(^{3}\)/s), \( C \) = runoff coefficient, \( i \) = intensity (in/hr), \( A \) = area (ac).
Note: When \( C \) is dimensionless, \( i \) in in/hr, and \( A \) in acres, \( Q \) comes out directly in ft\(^{3}\)/s (because 1 ac·in/hr ≈ 1.008 ft\(^{3}\)/s ≈ 1 cfs).
Step 3 – Calculate the weighted runoff coefficient:
\( C_{\text{weighted}} = \frac{C_1 A_1 + C_2 A_2}{A_{\text{total}}} = \frac{(0.30)(3) + (0.70)(5)}{8} = \frac{0.90 + 3.50}{8} = \frac{4.40}{8} = 0.55 \)
Step 4 – Compute peak runoff:
\( Q = C_{\text{weighted}} \cdot i \cdot A_{\text{total}} = 0.55 \times 4.0 \times 8 = 17.6 \approx 18 \;\text{ft}^3\text{/s} \)
Correct answer: (C)
Problem 2
A \( 50\,\mathrm{ac} \) watershed consists of \( 20\,\mathrm{ac} \) pasture (\( C = 0.15 \)), \( 10\,\mathrm{ac} \) commercial (\( C = 0.90 \)), and \( 20\,\mathrm{ac} \) woodland (\( C = 0.25 \)). If time of concentration is \( 25\,\mathrm{min} \) and the corresponding intensity is \( 7\,\mathrm{in/hr} \), what is the peak runoff?
A) \( 65\,\mathrm{ft}^3/\mathrm{s} \)
B) \( 120\,\mathrm{ft}^3/\mathrm{s} \)
C) \( 180\,\mathrm{ft}^3/\mathrm{s} \)
D) \( 240\,\mathrm{ft}^3/\mathrm{s} \)
Answer: B) \( 120\,\mathrm{ft}^3/\mathrm{s} \)
Step 1 – Identify given information:
| Land Use | Area (ac) | C |
|---|---|---|
| Pasture | 20 | 0.15 |
| Commercial | 10 | 0.90 |
| Woodland | 20 | 0.25 |
| Total | 50 | -- |
Time of concentration \( t_c = 25 \) min, corresponding intensity \( i = 7 \) in/hr.
Step 2 – Weighted runoff coefficient (Rational Method):
\( C_w = \frac{\sum C_i A_i}{A_{\text{total}}} \)
Step 3 – Solve:
\( C_w = \frac{(20)(0.15) + (10)(0.90) + (20)(0.25)}{50} = \frac{3.0 + 9.0 + 5.0}{50} = \frac{17.0}{50} = 0.34 \)
\( Q = C_w \cdot i \cdot A = 0.34 \times 7 \times 50 = 119 \;\text{ft}^3\text{/s} \)
Step 4 – Answer: \( Q \approx 120 \) cfs. The commercial area (C = 0.90) dominates the runoff despite being the smallest area. Option (A) 65 cfs would require a much lower weighted C; options (C) and (D) overestimate the impervious contribution.
Correct answer: (B)
Problem 3
A watershed has \( CN = 92 \) and receives rainfall depth \( P = 2.0\,\mathrm{in} \). What is the direct runoff depth \( Q \) (SCS CN method)?
A) \( 0.6\,\mathrm{in} \)
B) \( 1.2\,\mathrm{in} \)
C) \( 1.7\,\mathrm{in} \)
D) \( 2.0\,\mathrm{in} \)
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 4
An IDF curve (to be sketched) passes through (\( 10\,\mathrm{min} \), \( 12\,\mathrm{in/hr} \)), (\( 30\,\mathrm{min} \), \( 7\,\mathrm{in/hr} \)), and (\( 60\,\mathrm{min} \), \( 4\,\mathrm{in/hr} \)). Estimate the intensity for \( t_c = 40\,\mathrm{min} \) by linear interpolation, then compute \( Q \) for \( C = 0.50 \) and \( A = 15\,\mathrm{ac} \) (rational method).
A) \( 30\,\mathrm{ft}^3/\mathrm{s} \)
B) \( 45\,\mathrm{ft}^3/\mathrm{s} \)
C) \( 60\,\mathrm{ft}^3/\mathrm{s} \)
D) \( 75\,\mathrm{ft}^3/\mathrm{s} \)
Answer: B) \( 45\,\mathrm{ft}^3/\mathrm{s} \)
Step 1 – Identify given information:
IDF data points: \( (10\text{ min},\;12\text{ in/hr}) \), \( (30\text{ min},\;7\text{ in/hr}) \), \( (60\text{ min},\;4\text{ in/hr}) \). Also: \( t_c = 40 \) min, \( C = 0.50 \), \( A = 15 \) ac.
Step 2 – Linear interpolation for intensity:
Between the 30-min and 60-min data points:
\( i(40) = 7 + \frac{4 - 7}{60 - 30}(40 - 30) = 7 + \frac{-3}{30}(10) = 7 - 1.0 = 6.0 \;\text{in/hr} \)
Step 3 – Apply the Rational Method:
\( Q = C \cdot i \cdot A = 0.50 \times 6.0 \times 15 = 45 \;\text{ft}^3\text{/s} \)
Step 4 – Answer: \( Q = 45 \) cfs. The IDF curve shows that intensity decreases with increasing storm duration. Using the correct \( t_c \) to enter the IDF curve is essential for accurate peak-flow estimates.
Correct answer: (B)
Problem 5
A \( 400\,\mathrm{ft} \times 250\,\mathrm{ft} \) parking lot has runoff coefficient \( C = 0.40 \). If the rainfall intensity is \( 6.0\,\mathrm{in/hr} \), what is the peak runoff \( Q \) (rational method)?
A) \( 3.7\,\mathrm{ft}^3/\mathrm{s} \)
B) \( 5.5\,\mathrm{ft}^3/\mathrm{s} \)
C) \( 7.9\,\mathrm{ft}^3/\mathrm{s} \)
D) \( 11\,\mathrm{ft}^3/\mathrm{s} \)
Answer: B) \( 5.5\,\mathrm{ft}^3/\mathrm{s} \)
Step 1 – Identify given information:
| Parameter | Value |
|---|---|
| Lot dimensions | \( 400 \times 250 \) ft |
| Runoff coefficient \( C \) | 0.40 |
| Rainfall intensity \( i \) | \( 6.0 \) in/hr |
Step 2 – Rational Method formula:
\( Q = C \cdot i \cdot A \)
where \( Q \) is peak runoff (cfs), \( C \) is the runoff coefficient, \( i \) is rainfall intensity (in/hr), and \( A \) is drainage area (ac). Note: 1 ac = 43,560 ft\(^2\).
Step 3 – Convert area and solve:
\( A = \frac{400 \times 250}{43{,}560} = \frac{100{,}000}{43{,}560} = 2.296 \;\text{ac} \)
\( Q = 0.40 \times 6.0 \times 2.296 = 5.51 \;\text{ft}^3\text{/s} \)
Step 4 – Answer: \( Q \approx 5.5 \) ft\(^3\)/s. Option (A) 3.7 would result from a smaller area or lower \( C \); options (C) and (D) are too large for this small lot.
Correct answer: (B)
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 8 to 12 questions for Water Resources and Environmental Engineering out of the 110 on the exam, roughly 7 to 11 percent. 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
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.