FE Civil · Chapter 11 · 14–21 exam questions

FE Civil Water Resources & Env.

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

What the FE tests in Water Resources & Env.

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.

Key Water Resources & Env. formulas

  • Q=KnARH2/3S1/2Q = \frac{K}{n}AR_H^{2/3}S^{1/2}
    Manning's EquationFE Handbook p. 297
  • v=k1CRH0.63Sv0.54v = k_1 C R_H^{0.63} S_v^{0.54}
    Hazen-WilliamsFE Handbook p. 297
  • E=αv22g+yE = \frac{\alpha v^2}{2g} + y
    Specific EnergyFE Handbook p. 295
  • yc=(q2/g)1/3y_c = (q^2/g)^{1/3}
    Critical Depth (Rectangular)FE Handbook p. 296
  • Q=CIAQ = CIA
    Rational MethodFE Handbook p. 290
  • Q=(P0.2S)2P+0.8SQ = \frac{(P - 0.2S)^2}{P + 0.8S}
    SCS Runoff EquationFE Handbook p. 290
  • Q=KAdhdxQ = -KA\frac{dh}{dx}
    Darcy's LawFE Handbook p. 292
  • Q=πK(h22h12)ln(r2/r1)Q = \frac{\pi K(h_2^2 - h_1^2)}{\ln(r_2/r_1)}
    Dupuit's FormulaFE Handbook p. 292
  • Q=2πT(h2h1)ln(r2/r1)Q = \frac{2\pi T(h_2 - h_1)}{\ln(r_2/r_1)}
    Thiem EquationFE Handbook p. 293
  • BODt=L0(1ekt)BOD_t = L_0(1-e^{-kt})
    BOD DecayFE Handbook p. 321
  • vo=Q/Asurfacev_o = Q/A_{surface}
    Clarifier Overflow RateFE Handbook p. 339
  • vs=Q/Aplanv_s = Q/A_{plan}
    Filtration RateFE Handbook p. 341
  • CT=C×t10CT = C \times t_{10}
    Disinfection CTFE Handbook p. 346
  • W˙=γQH/η\dot W = \gamma Q H / \eta
    Pump Power EquationFE Handbook p. 191
  • NPSHA=Hpa+HshLHvpNPSH_A = H_{pa} + H_s - \sum h_L - H_{vp}
    NPSH AvailableFE Handbook p. 191
  • Hardness as CaCO3=Ci50EWi\text{Hardness as CaCO}_3 = \sum C_i \tfrac{50}{EW_i}
    Total Hardness as CaCO₃

Sample Water Resources & Env. problems

Q1. Manning's equation uses the factor K=1.486K = 1.486 in US Customary units and K=1.0K = 1.0 in SI units. What happens to the computed discharge if an engineer accidentally uses K=1.0K = 1.0 with US Customary inputs?

Answer: The discharge is underestimated by about 33%

Explain it simply

If you use K=1.0K = 1.0 instead of 1.486, you get Qwrong=Qcorrect/1.486Q_{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.

Q2. A trapezoidal earth channel (n=0.022n = 0.022) has a bottom width of b=3mb = 3\,\text{m}, side slopes of 2H:1V, and a flow depth of y=1.5my = 1.5\,\text{m} on a slope of S=0.0004S = 0.0004. What is the discharge?

Answer: 7.8m3/s7.8\,\text{m}^3/\text{s}

Explain it simply

For a trapezoid with 2:1 side slopes: A=(b+zy)y=(3+2×1.5)(1.5)=6×1.5=9.0A = (b + zy)y = (3 + 2 \times 1.5)(1.5) = 6 \times 1.5 = 9.0 m2^2. P=b+2y1+z2=3+2(1.5)1+4=3+35=3+6.708=9.708P = b + 2y\sqrt{1 + z^2} = 3 + 2(1.5)\sqrt{1 + 4} = 3 + 3\sqrt{5} = 3 + 6.708 = 9.708 m. RH=9.0/9.708=0.927R_H = 9.0/9.708 = 0.927 m. SI units so K=1.0K = 1.0. Q=(1/0.022)(9.0)(0.927)2/3(0.0004)1/2=45.45×9.0×0.951×0.02=7.8Q = (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 m3^3/s. Choice A omits the side slopes from the area. Choice B uses K=1.486K = 1.486 (wrong for SI). Choice D uses the bottom width alone for the wetted perimeter.

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

Common Water Resources & Env. mistakes on the FE

  • 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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