Hydrology, open-channel and pipe flow, pumps, stormwater, groundwater, and water quality and treatment.
NCEES lists these Water Resources and Environmental Engineering topics for the FE Civil: basic hydrology: infiltration, rainfall, runoff, watersheds; basic hydraulics: Manning, Bernoulli, open-channel and pipe flow; pumps and water distribution systems; flood control and stormwater: hydrographs, routing, detention; collection systems and groundwater: wells and drawdown; water quality, testing and standards; water and wastewater treatment.
Choose any section below to get a free FE Civil practice problem. Work it with the FE Reference Handbook open, tap your answer, then check the solution.
The rational method gives peak runoff for small watersheds: Q = CiA, where C is the runoff coefficient, i the rainfall intensity for a storm as long as the time of concentration, and A the area.
A 12 ha watershed is 8 ha of grass (C = 0.30) and 4 ha of pavement (C = 0.80). Overland flow to the channel takes 15 min, and the runoff then travels 1,200 m along the channel at 1.0 m/s. The local IDF curve is i = 2,500/(t + 15) mm/h, with t in minutes. What is the peak flow by the rational method?
C = [8(0.30) + 4(0.80)]/12 = 0.467. Channel time = 1,200/1.0 = 1,200 s = 20 min, so tc = 15 + 20 = 35 min. i = 2,500/(35 + 15) = 50.0 mm/h. Q = CiA/360 = 0.467(50.0)(12)/360 = 0.78 m³/s.
Common trap: (B) Averages the two C values without weighting by area.
Rational MethodRunoff Curve NumberTime of ConcentrationThat was 1 of 4 problems in this section. The full course has the lesson, a worked example and every problem. Start your 3-day free trial →
Manning's equation (SI): V = (1/n) R2/3 S1/2, where R = A/P is the hydraulic radius (area over wetted perimeter) and S is the bed slope.
A trapezoidal channel has a 3.0 m bottom width and 2H:1V side slopes, and flows 1.2 m deep. With n = 0.015 and S = 0.0008, what is the discharge by Manning's equation?
A = (b + zy)y = (3.0 + 2·1.2)(1.2) = 6.48 m². P = b + 2y√(1 + z²) = 3.0 + 2(1.2)√5 = 8.37 m. R = 0.775 m. V = (1/0.015)(0.775)2/3(0.0008)1/2 = 1.590 m/s. Q = VA = 10.30 m³/s.
Common trap: (B) Uses the rectangular wetted perimeter b + 2y, ignoring the sloped sides.
ManningHydraulic RadiusFroude NumberCritical DepthSpecific EnergyThat was 1 of 4 problems in this section. The full course has the lesson, a worked example and every problem. Start your 3-day free trial →
Head loss in a pipe is the Darcy–Weisbach equation: hf = f(L/D)(V²/2g), with V = Q/A.
Water flows at 0.05 m³/s through a 200 mm pipe, 500 m long, with f = 0.02. What is the head loss?
V = Q/A = 0.05/(π·0.2²/4) = 1.59 m/s. hf = f(L/D)(V²/2g) = 0.02(2,500)(2.53)/19.62 = 6.5 m.
Common trap: (B) Uses V²/g instead of V²/2g.
Darcy-WeisbachHazen-WilliamsPump PowerEnergy EquationMinor LossesThat was 1 of 4 problems in this section. The full course has the lesson, a worked example and every problem. Start your 3-day free trial →
Development raises C and usually shortens the time of concentration, so both C and i rise and the peak runoff grows.
An 8 ha site has C = 0.30 and tc = 40 min before development, and C = 0.75 and tc = 20 min after. The design IDF curve is i = 3,000/(t + 20) mm/h, with t in minutes. By how much does the rational-method peak flow rise?
Pre: i = 3,000/(40 + 20) = 50.0 mm/h, Q = 0.30(50.0)(8)/360 = 0.333 m³/s. Post: i = 3,000/(20 + 20) = 75.0 mm/h, Q = 0.75(75.0)(8)/360 = 1.250 m³/s. Increase = 0.92 m³/s.
Common trap: (B) Keeps the pre-development intensity for both cases, ignoring the shorter tc.
Detention PondTime of ConcentrationHydrographThat was 1 of 4 problems in this section. The full course has the lesson, a worked example and every problem. Start your 3-day free trial →
Darcy's law applies underground: q = Ki, with seepage (pore) velocity v = Ki/n.
A well in an unconfined aquifer pumps steadily at 25 L/s. Two observation wells, 15 m and 60 m from the pumped well, show water-table heights of 18.2 m and 19.6 m above the impermeable base. What is the hydraulic conductivity?
Unconfined Thiem: Q = πK(h2² − h1²)/ln(r2/r1), so K = Q ln(r2/r1)/[π(h2² − h1²)]. h2² − h1² = 19.6² − 18.2² = 52.92 m². ln(60/15) = 1.386. K = 0.025(1.386)/[π(52.92)] = 2.08 × 10−4 m/s.
Common trap: (B) Uses (h2 − h1) instead of (h2² − h1²) for an unconfined aquifer.
DarcyWellsDrawdownTransmissivityThat was 1 of 4 problems in this section. The full course has the lesson, a worked example and every problem. Start your 3-day free trial →
Biochemical oxygen demand: BODt = L0(1 − e−kt), where L0 is the ultimate BOD and k the rate constant (base e, per day).
A wastewater has a 5-day BOD of 180 mg/L at 20°C, with k = 0.23/day (base e). Using θ = 1.047, what BOD is exerted in 10 days at 25°C?
L0 = BOD5/(1 − e−0.23(5)) = 180/0.683 = 263.4 mg/L. k25 = 0.23(1.047)5 = 0.289/day. BOD10 = 263.4(1 − e−0.289(10)) = 249 mg/L.
Common trap: (B) Skips the temperature correction and uses k = 0.23/day at 25°C.
BODSedimentationDisinfectionMass BalanceThat was 1 of 4 problems in this section. The full course has the lesson, a worked example and every problem. Start your 3-day free trial →
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