FE Civil · Chapter 11 · 14–21 questions

FE Civil Water Resources & Env.

Q: How many FE Civil Water Resources & Env. questions are on the exam?

NCEES includes roughly 14–21 Water Resources & Env. questions on the computer-based FE Civil exam, which has 110 questions total over about six hours.

Q: What Water Resources & Env. topics does the FE Civil exam cover?

The FE Civil exam tests Water Resources & Env. across: Hydraulics, Hydrology & Groundwater, Water Quality & Treatment.

Q: Is FE Civil Water Resources & Env. hard?

Water Resources & Env. is very learnable for the FE. It rewards recognizing a handful of standard problem types and applying the right FE Reference Handbook formula — not deep theory. Practicing past-style problems is the fastest way to master it.

Q: Where can I find free FE Civil Water Resources & Env. practice problems?

FE for Raccoons has free Water Resources & Env. lessons and practice problems with step-by-step explanations — part of a 1,126-problem bank covering all 15 FE Civil topics, free at fe4raccoons.com.

This is the highest-weighted chapter — covering open channel flow, hydrology, groundwater, and environmental engineering.

Practice Water Resources & Env. free Full exam guide

What the FE tests

The Water Resources & Env. skills NCEES checks

Hydraulics

As a civil engineer, you design open channels, storm sewers, and water distribution systems using Manning's equation, Hazen-Williams, and weir formulas. You classify flow as subcritical or supercritical using the Froude number, compute critical depth for channel transitions, and size pumps and pipes to deliver adequate pressure and flow while guarding against cavitation.

Hydrology & Groundwater

As a civil engineer, you estimate peak runoff using the Rational Method and SCS/NRCS curve number, predict hydrograph shapes with unit hydrograph theory, and evaluate groundwater flow with Darcy's law and well hydraulics. These methods drive the design of storm drains, detention ponds, dewatering systems, and water supply wells.

Water Quality & Treatment

As a civil engineer, you evaluate water and wastewater treatment performance using BOD, dissolved oxygen models, and design parameters like overflow rate, F:M ratio, and solids residence time. You size clarifiers, filters, activated sludge basins, and disinfection systems; set chlorine doses and CT; and check hardness and contaminant levels against Safe Drinking Water Act and NPDES limits.

Reference

Key Water Resources & Env. formulas

$Q = \frac{K}{n}AR_H^{2/3}S^{1/2}$ Manning's EquationFE Handbook p. 297
$v = k_1 C R_H^{0.63} S_v^{0.54}$ Hazen-WilliamsFE Handbook p. 297
$E = \frac{\alpha v^2}{2g} + y$ Specific EnergyFE Handbook p. 295
$y_c = (q^2/g)^{1/3}$ Critical Depth (Rectangular)FE Handbook p. 296
$Q = CIA$ Rational MethodFE Handbook p. 290
$Q = \frac{(P - 0.2S)^2}{P + 0.8S}$ SCS Runoff EquationFE Handbook p. 290
$Q = -KA\frac{dh}{dx}$ Darcy's LawFE Handbook p. 292
$Q = \frac{\pi K(h_2^2 - h_1^2)}{\ln(r_2/r_1)}$ Dupuit's FormulaFE Handbook p. 292
$Q = \frac{2\pi T(h_2 - h_1)}{\ln(r_2/r_1)}$ Thiem EquationFE Handbook p. 293
$BOD_t = L_0(1-e^{-kt})$ BOD DecayFE Handbook p. 321
$v_o = Q/A_{surface}$ Clarifier Overflow RateFE Handbook p. 339
$v_s = Q/A_{plan}$ Filtration RateFE Handbook p. 341
$CT = C \times t_{10}$ Disinfection CTFE Handbook p. 346
$\dot W = \gamma Q H / \eta$ Pump Power EquationFE Handbook p. 191
$NPSH_A = H_{pa} + H_s - \sum h_L - H_{vp}$ NPSH AvailableFE Handbook p. 191
$\text{Hardness as CaCO}_3 = \sum C_i \tfrac{50}{EW_i}$ Total Hardness as CaCO₃

Try it

Sample Water Resources & Env. problems

Q1Manning's equation uses the factor $K = 1.486$ in US Customary units and $K = 1.0$ in SI units. What happens to the computed discharge if an engineer accidentally uses $K = 1.0$ with US Customary inputs?

The discharge is underestimated by about 33%

Explain it simply

If you use $K = 1.0$ instead of 1.486, you get $Q_{wrong} = Q_{correct}/1.486$ , which means your answer is about 67% of the correct value. That is an underestimate of about 33%. This is a classic FE trap because forgetting the conversion factor does not produce an obviously wrong answer. Choice A reverses the direction of the error. Choice C ignores the factor entirely. Choice D confuses the percentage.

Q2A trapezoidal earth channel ( $n = 0.022$ ) has a bottom width of $b = 3\,\text{m}$ , side slopes of 2H:1V, and a flow depth of $y = 1.5\,\text{m}$ on a slope of $S = 0.0004$ . What is the discharge?

$7.8\,\text{m}^3/\text{s}$

Explain it simply

For a trapezoid with 2:1 side slopes: $A = (b + zy)y = (3 + 2 \times 1.5)(1.5) = 6 \times 1.5 = 9.0$ m $^2$ . $P = b + 2y\sqrt{1 + z^2} = 3 + 2(1.5)\sqrt{1 + 4} = 3 + 3\sqrt{5} = 3 + 6.708 = 9.708$ m. $R_H = 9.0/9.708 = 0.927$ m. SI units so $K = 1.0$ . $Q = (1/0.022)(9.0)(0.927)^{2/3}(0.0004)^{1/2} = 45.45 \times 9.0 \times 0.951 \times 0.02 = 7.8$ m $^3$ /s. Choice A omits the side slopes from the area. Choice B uses $K = 1.486$ (wrong for SI). Choice D uses the bottom width alone for the wetted perimeter.

2 of 1,126 problems across all 15 chapters — the full bank, lessons, mastery tracking, and timed exam simulation live inside the app.

Avoid these

Common Water Resources & Env. mistakes

Manning's n is a roughness coefficient — in US Customary, multiply by 1.486 (K factor).
Hydraulic radius R_H = A/P is NOT the physical radius — for a full pipe, R_H = D/4.
Hazen-Williams C is the OPPOSITE convention from Manning's n — higher C means smoother pipe.
The Rational Method Q = CIA only works for small watersheds (< 200 acres).
SCS runoff equation gives depth in inches, NOT discharge in cfs.
Darcy velocity is NOT the actual seepage velocity — divide by porosity for real velocity.
Dupuit (unconfined) uses h² differences; Thiem (confined) uses linear h differences.
BOD₅ is the 5-day BOD, not the ultimate BOD (L₀) — they are different values.

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