Water Resources and Environmental Engineering ties with Geotechnical for the longest topic list in the specification, eleven items and 10–15 questions from rainfall to the treatment plant. This page takes the infrastructure side: runoff, channels and pipes, pumps, distribution networks, stormwater and sewers, and the civil fluid mechanics behind them.
Basic hydraulics: the Manning equation, the Bernoulli theorem, open-channel flow.
Pumps.
Water distribution systems.
Flood control: dams, routing, spillways.
Stormwater: detention, routing, quality.
Collection systems for wastewater and stormwater.
Groundwater: flow, wells, drawdown.
Water quality, ground and surface, and basic water chemistry.
Testing and standards for water, wastewater, air and noise.
Water and wastewater treatment, from biological processes to softening.
18 water resources problems
0 / 18 answered · 0 correct
Question 1 of 18 · Fluid Mechanics · 6–9 items
A rectangular water tank is 3 m wide and 5 m long and holds water 4 m deep. What is the resultant hydrostatic force on one 3-m-wide vertical end wall?
Given
End wall width b = 3 m
Water depth h = 4 m
γw = 9.81 kN/m³
Find
Resultant hydrostatic force on the end wall F (kN)
Solution
F = γhcA = 9.81 kN/m³ × (4/2) m × (3 m × 4 m) = 235.44 kN, acting at 2h/3 = 2.67 m below the surface
FE Reference Handbook, Fluid Mechanics — forces on submerged plane surfaces.
588.60 kN is the force on the 3 × 5 m tank floor (γh × floor area), not on the end wall.
470.88 kN uses the pressure at the bottom, γh, over the whole area; the average pressure is half of that.
392.40 kN uses the 5-m side wall; the end wall is the 3-m one.
F = γ(h/2)(bh) = 9.81 × 2 × (3 × 4) = 235.44 kN, the pressure at the centroid times the wetted area.
Question 2 of 18 · Fluid Mechanics · 6–9 items
A free-surface open channel has a rectangular cross-section with a width of 5 m and a flow depth of 2 m. Using Manning's equation with n = 0.015 and a channel slope of 0.001, what is the average velocity of flow?
Given
Channel width b = 5 m
Flow depth y = 2 m
Manning n = 0.015
Slope S = 0.001
Find
Average velocity V (m/s)
Solution
R = A/P = (5 × 2)/(5 + 2 × 2) = 1.111 m
V = (1/n)R2/3S1/2 = (1/0.015)(1.111)2/3(0.001)1/2 = 2.26 m/s
FE Reference Handbook, Fluid Mechanics / Water Resources — Manning's equation (SI, K = 1.0).
1.68 m/s uses P = 2(b + y) = 14 m, the full perimeter of the rectangle; the free surface is not wetted.
A = 10 m², P = 5 + 2 × 2 = 9 m, R = 1.111 m, so V = (1/0.015)(1.111)2/3(0.001)1/2 = 2.26 m/s.
2.67 m/s counts only one side wall (P = b + y = 7 m), which overstates the hydraulic radius.
3.35 m/s takes R equal to the flow depth (2 m); that approximation holds only for very wide channels.
Question 3 of 18 · Fluid Mechanics · 6–9 items
A circular storm sewer flows exactly half full. Compared with the same pipe flowing full on the same slope and with the same Manning n, the half-full pipe has:
Solution
A = πD²/8; P = πD/2; R = A/P = D/4 = Rfull
V ∝ R2/3S1/2/n → Vhalf = Vfull; Q = VA → Qhalf = Qfull/2
FE Reference Handbook, Water Resources — partially full circular pipes.
At half depth A = πD²/8 and P = πD/2, so R = D/4, the same as full; Manning gives the same V, and with half the area Q is half.
Velocity depends on R, not on depth; R at half depth equals the full-pipe value, so V does not halve.
The velocity is the same, but the flow area is half, so the discharge is half too.
V and Q exceed full-pipe values only at depths of roughly 0.8D to 0.95D, not at half depth.
Question 4 of 18 · Water Resources and Environmental Engineering · 10–15 items
An engineer proposes using the Hazen-Williams equation to design a piping system for hot water at 85°C. Why is the Darcy-Weisbach equation preferred over Hazen-Williams for this application?
Solution
hf = f(L/D)(V²/2g), f = f(Re, ε/D), Re = VD/ν
ν of water ≈ 1.0 × 10−6 m²/s at 20 °C vs about 0.34 × 10−6 m²/s at 85 °C. FE Reference Handbook, Fluid Mechanics — head loss in pipes.
Hot water has lower viscosity, which raises Re, so laminar flow becomes less likely, not certain.
Darcy–Weisbach does need Re, through the friction factor from the Moody diagram; that is exactly how it captures viscosity.
Thermal growth of the diameter is negligible; it is not why Hazen–Williams fails for hot water.
The Hazen–Williams C is set by pipe material and age, not by water temperature.
Hazen–Williams is empirical for water near 60 °F; at 85 °C the viscosity is about one-third of that, and only Darcy–Weisbach accounts for it through Re.
Question 5 of 18 · Water Resources and Environmental Engineering · 10–15 items
Two pipes, A and B, have identical length, diameter and head loss. Pipe A has a Hazen–Williams coefficient C = 130 and Pipe B has C = 100. How much more flow can Pipe A carry?
Given
Pipe A: C = 130
Pipe B: C = 100
Same L, D and head loss
Find
Additional flow of Pipe A relative to B (%)
Solution
Q = 0.432CD2.63S0.54 (US) → Q ∝ C at fixed D and S
QA/QB = 130/100 = 1.30
FE Reference Handbook, Fluid Mechanics — Hazen–Williams equation.
15 % is (1.3)0.54 − 1; the 0.54 exponent applies to the slope term, while Q is linear in C.
23 % is (130 − 100)/130, the difference taken relative to Pipe A instead of Pipe B.
With D and the head-loss slope equal, Q ∝ C, so QA/QB = 130/100 = 1.30, or 30 % more.
63 % is (1.3)1.852 − 1, the ratio of head losses at equal flow, not of flows at equal head loss.
69 % is (1.3)² − 1; Q does not vary with C squared.
Question 6 of 18 · Water Resources and Environmental Engineering · 10–15 items
Hydraulic analysis for a proposed highway culvert has produced the headwater (HW) elevations below for the design discharge. Based on these results, what is the governing design headwater elevation?
Item
Elevation
Inlet invert
100.00 ft
Inlet-control HW
104.55 ft
Outlet-control HW
106.20 ft
Solution
Governing HW = max(HWinlet, HWoutlet) = max(104.55, 106.20) = 106.20 ft (FHWA HDS-5 procedure). FE Reference Handbook, Water Resources — culverts.
6.20 ft is the outlet-control headwater measured as a depth above the 100.00-ft invert; the question asks for an elevation.
104.55 ft is the inlet-control headwater; taking the lower value ignores that outlet control needs more head.
105.38 ft averages the two; a culvert operates under whichever control demands the higher headwater.
The culvert works under the control that requires the higher headwater, here outlet control, so the design HW is 106.20 ft.
Question 7 of 18 · Water Resources and Environmental Engineering · 10–15 items
A drainage channel has a length of 2,000 ft and a slope of 1.0%. Using the Kirpich equation Tc = 0.0078 L0.77 S−0.385 (where Tc is in minutes, L in ft, and S in ft/ft), determine the design rainfall intensity using the IDF equation i = 45/(t+10).
FE Reference Handbook, Water Resources — time of concentration and rainfall intensity.
0.9 in/hr comes from entering S = 0.001 (a decimal slip), which inflates Tc to about 39 min.
Tc = 0.0078(2000)0.77(0.010)−0.385 = 16.0 min, so i = 45/(16.0 + 10) = 1.7 in/hr.
2.8 in/hr drops the +10 in the IDF denominator (45/16.0).
3.5 in/hr enters S = 1.0 instead of 0.010 ft/ft, giving Tc ≈ 2.7 min.
4.5 in/hr is 45/10, the IDF evaluated at zero duration instead of at Tc.
Question 8 of 18 · Water Resources and Environmental Engineering · 10–15 items
Using the IDF data below, estimate the rainfall intensity for a 20-minute duration by fitting the power-law relation i = a·tb through the two points (log-log interpolation).
Duration (min)
Intensity (in/hr)
15
4.8
30
3.2
Given
15 min intensity = 4.8 in/h
30 min intensity = 3.2 in/h
Duration = 20 min
Find
Rainfall intensity (in/h)
Solution
b = ln(3.2/4.8)/ln(30/15) = −0.585
i20 = 4.8 × (20/15)−0.585 = 4.06 in/hr
FE Reference Handbook, Water Resources — intensity-duration-frequency.
b = ln(3.2/4.8)/ln(30/15) = −0.585, so i = 4.8 × (20/15)−0.585 = 4.06 in/hr.
4.14 in/hr is semi-log interpolation (intensity linear in log t), not the log-log power law the stem asks for.
4.27 in/hr is straight linear interpolation, which overestimates on a curve that is concave in linear axes.
4.80 in/hr is the 15-min intensity read straight from the table, without interpolating to 20 min.
5.68 in/hr flips the sign of b, so intensity would rise with duration.
Question 9 of 18 · Water Resources and Environmental Engineering · 10–15 items
A heterogeneous watershed has area 320 acres, mixed residential and commercial land use (C ranges 0.4–0.9), and time of concentration 35 minutes. Evaluate whether the Rational method is appropriate for peak discharge estimation.
Solution
Rational method Q = CiA assumes uniform intensity over the area for a duration ≥ Tc; agencies typically limit it to about 200 acres. FE Reference Handbook, Water Resources — rational method.
A 35-min Tc is not by itself a disqualifier; the size and heterogeneity are.
A weighted C handles mixed cover only within the method's size limit; it does not fix the 320-acre area.
The method assumes uniform rainfall over the whole area for the full Tc, which breaks down on large watersheds.
Tc in range does not override the area limit.
Most agencies cap the Rational method near 200 acres; at 320 acres with C from 0.4 to 0.9, a hydrograph method on subareas is the defensible choice.
Question 10 of 18 · Water Resources and Environmental Engineering · 10–15 items
A water tank keeps the hydraulic grade line at elevation 620 ft. Three demand nodes have the elevations and total head losses (major + minor) from the tank shown in the Given list. Rank the nodes from lowest to highest residual pressure.
Pressure = 0.433 psi per ft of head. FE Reference Handbook, Fluid Mechanics — energy equation.
A, B, C lists the nodes from highest to lowest pressure, the reverse of what was asked.
C, A, B ranks by static head (620 − elevation) alone; subtracting the head losses moves B below A.
B, C, A orders by head loss from largest to smallest, again leaving out elevation.
A, C, B orders the nodes by head loss alone (45, 50, 62 ft) and ignores elevation.
Pressure head = HGL − loss − elevation: C = 620 − 50 − 560 = 10 ft, B = 620 − 62 − 540 = 18 ft, A = 620 − 45 − 550 = 25 ft.
Question 11 of 18 · Water Resources and Environmental Engineering · 10–15 items
A small urban watershed (area 15 acres, composite runoff coefficient C = 0.65, time of concentration 12 min) is designed for a 25-year storm. The IDF curve gives i = 5.0 in/hr at 12 min. Is the rational method appropriate, and what is the design peak discharge?
FE Reference Handbook, Water Resources — rational method.
15 acres is far below the usual 200-acre ceiling, so size is not a reason to reject it.
75 cfs is i × A without the runoff coefficient, as if the basin were fully impervious.
A 15-acre urban basin is well inside the method's range, and Q = CiA = 0.65 × 5.0 × 15 = 48.8 cfs (49.1 with the 1.008 unit factor).
The return period enters through the IDF intensity; the rational method handles a 25-year design storm.
4.9 cfs is a factor of 10 low, a decimal slip in the product CiA.
Question 12 of 18 · Water Resources and Environmental Engineering · 10–15 items
When analyzing the effects of climate change on watershed hydrology, which parameter is most sensitive in altering the design criteria of stormwater management systems?
Solution
Peak runoff Q = CiA and detention volumes scale with design intensity i from IDF data, so nonstationary rainfall statistics change design criteria most.
Design storms come from IDF curves; shifts in extreme-rainfall frequency and intensity change peak flows and required capacity directly.
Watershed area is fixed by topography; climate change does not alter it in any design-relevant way.
Vegetation affects runoff coefficients modestly; the design storm itself is the dominant input.
Soil properties change on geologic time scales, too slowly to drive stormwater design criteria.
Question 13 of 18 · Water Resources and Environmental Engineering · 10–15 items
An IDF table lists 10-year intensities: 5.8 in/hr at 10 min, 4.2 in/hr at 15 min, 3.1 in/hr at 20 min. Interpolate linearly to find the intensity for 12-minute duration, then convert to total rainfall depth for a 12-minute storm.
FE Reference Handbook, Water Resources — IDF curves.
61.92 in multiplies the intensity by 12 minutes without converting minutes to hours.
5.16 is the interpolated intensity (in/hr) reported as a depth; it still needs × (12/60) hr.
1.29 in multiplies the interpolated 5.16 in/hr by 15 min instead of 12 min (5.16 × 15/60).
1.16 in uses the 10-min intensity (5.8 in/hr) without interpolating.
i12 = 5.8 + (4.2 − 5.8)(2/5) = 5.16 in/hr and depth = 5.16 × 12/60 = 1.03 in.
Question 14 of 18 · Water Resources and Environmental Engineering · 10–15 items
Standard Diameter (in)
18
21
24
27
30
A storm sewer must convey 18 cfs at slope S = 0.006 with Manning n = 0.013 while flowing 75 % full (d/D = 0.75). Using Manning's equation, what is the smallest standard pipe diameter from the list that carries the flow, and does it keep at least 2.5 ft/s?
Given
Design flow Q = 18 ft3/s
Slope S = 0.006
Manning roughness n = 0.013
Flow depth ratio d/D = 0.75
Minimum velocity = 2.5 ft/s
Standard diameters = 18, 21, 24, 27, 30 in
Find
Minimum standard pipe diameter (in)
Solution
θ = 2 arccos(1 − 2d/D); A = (D²/8)(θ − sin θ); P = Dθ/2
Q = (1.486/n)AR2/3S1/2: 24 in → 16.0 cfs, 27 in → 21.9 cfs
V = 18/A27 = 5.6 ft/s ≥ 2.5 ft/s
FE Reference Handbook, Water Resources — Manning's equation, partially full circular pipes.
An 18-in pipe carries only about 7.4 cfs at 75 % depth, well short of 18 cfs.
A 21-in pipe carries about 11.2 cfs at 75 % depth, still short of 18 cfs.
A 24-in pipe carries about 16.0 cfs at 75 % depth (17.5 cfs even flowing full), short of 18 cfs; choosing it skips the capacity check.
At 75 % depth a 27-in pipe carries about 21.9 cfs ≥ 18 cfs, with V ≈ 5.6 ft/s > 2.5 ft/s.
A 30-in pipe works (≈ 29 cfs) but is not the smallest size that does.
Question 15 of 18 · Water Resources and Environmental Engineering · 10–15 items
For a water distribution system, how does selecting pumps with lower specific energy consumption impact long-term operational sustainability?
Solution
P = γQH/η; energy per volume = γH/η
Life-cycle cost of pumps is dominated by energy. FE Reference Handbook, Fluid Mechanics — pump power.
Pumping energy is the largest operating cost of most distribution systems; lower kWh per unit volume cuts both cost and emissions over the life of the station.
Over a 20- to 30-year life, small efficiency differences compound into large energy totals.
Specific energy is an efficiency measure, not a capacity rating; it acts on every hour of operation.
Lower specific energy consumption is by definition an energy-efficiency gain.
Question 16 of 18 · Water Resources and Environmental Engineering · 10–15 items
When designing a detention basin for urban flood control, how does the determination of time of concentration influence the basin storage requirements and design hydrograph?
Solution
Storage ≈ volume of the inflow hydrograph above the allowable release; both the peak (from i at Tc) and the time base depend on Tc. FE Reference Handbook, Water Resources — hydrographs and detention.
Tc is a time, not an area; it sets the design intensity and the hydrograph timing.
The inflow hydrograph shape changes the volume above the allowable outflow, so storage is affected.
Tc fixes the critical storm duration and the shape and peak of the inflow hydrograph, which together set the storage needed to hold outflow to the target.
The design intensity itself is read from the IDF curve at a duration tied to Tc.
Question 17 of 18 · Water Resources and Environmental Engineering · 10–15 items
A rectangular channel (bottom width 8.0 ft, bed slope 0.0020, Manning n = 0.025) carries 120 ft³/s. Find the normal depth and the Froude number at normal depth, and decide whether a hydraulic jump can form upstream of a tailwater depth of 4.2 ft. Use g = 32.2 ft/s².
Given
Bottom width b = 8.0 ft
Bed slope S = 0.0020
Manning n = 0.025
Discharge Q = 120 ft³/s
Tailwater depth yt = 4.2 ft
g = 32.2 ft/s²
Find
Normal depth yn, Froude number Fr and jump check
Solution
Q = (1.486/n)AR2/3S1/2, A = 8y, P = 8 + 2y → yn = 3.66 ft
yc = (q²/g)1/3 = (15²/32.2)1/3 = 1.91 ft < yn → mild slope, no jump
FE Reference Handbook, Fluid Mechanics / Water Resources — open-channel flow.
3.24 ft results from a wetted perimeter of b + y, counting only one wall.
1.91 ft is the critical depth yc = (q²/g)1/3, which by definition has Fr = 1; it is not the normal depth.
4.94 ft uses the SI Manning constant 1.0 with US units; in ft-s units it is 1.486.
Manning gives yn = 3.66 ft (V = 4.10 ft/s), Fr = 4.10/√(32.2 × 3.66) = 0.38; flow arriving subcritical cannot jump.
A jump needs supercritical approach flow (Fr > 1); tailwater above normal depth just backs water up.
Question 18 of 18 · Water Resources and Environmental Engineering · 10–15 items
A trapezoidal earthen channel with bottom width 8 ft and 2H:1V side slopes has Manning n = 0.035 and a bed slope of 0.0012. What is the normal depth when it carries 95 ft³/s?
Given
Bottom width b = 8 ft
Side slopes z = 2 (2H:1V)
Manning n = 0.035
Bed slope S = 0.0012
Discharge Q = 95 ft³/s
Find
Normal depth yn (ft)
Solution
A = (b + zy)y; P = b + 2y√(1 + z²); R = A/P
95 = (1.486/0.035)·A·R2/3·(0.0012)1/2 → yn ≈ 2.97 ft (A = 41.4 ft², R = 1.95 ft)
FE Reference Handbook, Water Resources — Manning's equation, trapezoidal channels.
1.44 ft is the critical depth (Q²T/(gA³) = 1), not the uniform-flow depth.
2.59 ft takes the wetted perimeter as b + 2y, ignoring the slope length √(1 + z²) of the sides.
Solving 95 = (1.486/0.035)AR2/3(0.0012)1/2 with A = (8 + 2y)y and P = 8 + 2y√5 gives yn ≈ 2.97 ft.
3.63 ft uses the SI Manning constant 1.0 instead of 1.486 for ft-s units.
4.80 ft treats the section as rectangular (A = 8y), dropping the side-slope area.
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Common traps
Rational method units. Q = C·i·A with i in in/hr and A in acres gives cfs directly, to within about 1%. In SI the same formula needs a conversion factor.
Water power vs brake power. Power delivered to the water is γ·Q·H; the pump shaft needs that divided by the pump efficiency, and the motor needs more again.
NPSH. Cavitation is avoided when the available NPSH, computed from the suction side of your system in absolute pressure, exceeds the required NPSH from the pump curve.
Pipes in series and in parallel. In series the flows are equal and head losses add; in parallel the head losses are equal and flows add.
Sources: FE Reference Handbook, Civil Engineering chapter; Chow, Open-Channel Hydraulics; Mays, Water Resources Engineering.
Rainfall at the wrong duration
The rational method takes its design intensity from the intensity-duration-frequency chart for a storm that lasts as long as the time of concentration of the drainage area. Read it at an hour, or at the storm length, and the intensity, and with it the peak flow, comes out too low for a small site.
Fluid mechanics on a civil site
The civil exam has a separate Fluid Mechanics section of 6–9 questions. Its civil scenarios are taught here: hydrostatic force on a gate or a wall, the thrust a bend or reducer puts on its anchor block, and flow measurement on site with weirs, flumes and Venturi meters. For force on a submerged surface, the resultant acts at the center of pressure, below the centroid, and its magnitude is the centroid pressure times the area.
The civil exam shares water with FE Environmental, and on this site each topic is taught once. Treatment unit processes and water chemistry are on water and wastewater; aquifer hydraulics on groundwater, soils and sediments; watershed budgets on surface water and hydrology. Here the question is always about sizing or checking something built: a sewer, a culvert, a pump, a pond, a network. Engineering-economics comparisons of pumps and pipes are on engineering economics.
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