Water Resources and Environmental Engineering
10–15 of 110 itemsNCEES draws between 10 and 15 of the 110 from this area on any legal paper. Its outline lists 11 sub-topics; the bank holds 59 verified templates against them, which expand to 1,721 questions.
Sub-topics, from the NCEES outline
- ABasic hydrology (e.g., infiltration, rainfall, runoff, watersheds)
- BBasic hydraulics (e.g., Manning equation, Bernoulli theorem, open-channel flow)
- CPumps
- DWater distribution systems
- EFlood control (e.g., dams, routing, spillways)
- FStormwater (e.g., detention, routing, quality)
- GCollection systems (e.g., wastewater, stormwater)
- HGroundwater (e.g., flow, wells, drawdown)
- IWater quality (e.g., ground and surface, basic water chemistry)
- JTesting and standards (e.g., water, wastewater, air, noise)
- KWater and wastewater treatment (e.g., biological processes, softening, drinking water treatment)
5 worked problems
Real questions from the bank, at the depth the exam asks them. Pick an answer first: the reasoning opens when you do, along with why each wrong answer is tempting.
Question 1 of 5
Water discharges from a reservoir through a short pipe whose outlet sits 27 ft below the free surface. Minor and friction losses total 1.5 ft. The velocity at the outlet is most nearly:
Reservoir discharging through a short pipe whose outlet sits 27 ft below the free surface. Show the answer and the reasoning
Why A
Given: outlet 27 ft below the free surface; losses 1.5 ft. Relation: energy from the surface to the outlet, h = V^2/2g + hL, so V = sqrt(2g(h - hL)). Substitute: V = sqrt(64.4 x (27 - 1.5)). Result: V = 40.5 ft/s.
Why B is tempting
Ignored the head loss. Only the net available head converts to velocity.
Why C is tempting
Added the head loss instead of subtracting it.
Why D is tempting
Never took the square root, so the result is a head times an acceleration.
Source FHWA HDS-4, Introduction to Highway Hydraulics — Section 2.2, energy in open channel flow
Question 2 of 5
A municipality is checking whether an existing gravity sewer can carry the flow from a proposed development. The line is 12 in in diameter and runs 700 ft between two manholes, with invert elevations of 225.00 ft at the upstream manhole and 216.50 ft at the downstream one. The pipe material has a roughness coefficient of 0.011, and the development is being evaluated against the 25-yr storm. Flowing full, the capacity of the line is most nearly:
Show the answer and the reasoning
Why D
Given: D = 12 in, n = 0.011, a 700-ft run between inverts 225.00 and 216.50 ft. The 25-yr storm sizes the demand, not the pipe. Relation: S is the invert drop over the run; flowing full, Q = (1.486/n) A R^(2/3) S^(1/2) with R = D/4. Substitute: S = 8.50/700 = 0.0121 ft/ft; A = 0.785 ft^2, R = 0.250 ft. Result: Q = 4.640 cfs.
Why A is tempting
Used S rather than its square root, which understates the capacity of any slope flatter than 1:1.
Why B is tempting
Put the radius into the area formula in place of the diameter, which quarters the flow area.
Why C is tempting
Used the SI coefficient 1.0. In US customary units Manning carries 1.486.
Source FHWA HEC-22, Urban Drainage Design Manual, 3rd ed. — Chapters 3 and 7, hydrology and storm drains
Question 3 of 5
A 49-acre catchment has a runoff coefficient of 0.75. For a design storm with an average intensity of 2 in/hr over the time of concentration, the peak discharge by the rational method is most nearly:
Show the answer and the reasoning
Why B
Given: C = 0.75, i = 2 in/hr over the time of concentration, A = 49 ac. Relation: peak runoff Q = CiA, in cfs with i in in/hr and A in acres. Substitute: Q = (0.75)(2)(49). Result: Q = 73.50 cfs. Check: no conversion is needed because one acre-inch per hour is 1.008 cfs, and no frequency factor applies when no long return period is named.
Why A is tempting
Used 1 - C, the fraction that infiltrates, where the fraction that runs off belongs.
Why C is tempting
Applied the 1.25 frequency factor reserved for 50- to 100-year storms. No return period is named, so C is used as given.
Why D is tempting
Left the runoff coefficient out. Only part of the rainfall becomes runoff.
Source FHWA HEC-22, Urban Drainage Design Manual, 3rd ed. — Section 3.2.1, the rational method
Question 4 of 5
A stream flowing at 310 cfs carries a background concentration of 8 mg/L. A discharge of 14 cfs at 110 mg/L enters the stream. Assuming complete mixing, the downstream concentration is most nearly:
Show the answer and the reasoning
Why B
Given: stream 310 cfs at 8 mg/L; discharge 14 cfs at 110 mg/L. Relation: mass balance, C = (Qs Cs + Qw Cw)/(Qs + Qw). Substitute: C = (310 x 8 + 14 x 110)/324 = 4020.0/324. Result: C = 12.4 mg/L. Check: the result sits nearer the stream concentration because the stream carries more flow.
Why A is tempting
Reported the background concentration, as if the discharge had no effect on the stream at all.
Why C is tempting
Divided by the stream flow alone.
Why D is tempting
Added the concentrations directly, which ignores dilution altogether.
Source FHWA HEC-22, Urban Drainage Design Manual, 3rd ed. — Chapters 3 and 7, hydrology and storm drains
Question 5 of 5
A well fully penetrates a confined aquifer 20 ft thick with a hydraulic conductivity of 50 ft/day. Under steady pumping the head difference between an observation point 200 ft away and the 1.0-ft diameter well is 4 ft. The well discharge is most nearly:
Well fully penetrating a confined aquifer 20 ft thick, with the piezometric surface drawn down toward the well and two observation radii on either side. Show the answer and the reasoning
Why C
Given: K = 50 ft/day, b = 20 ft, a 4-ft head difference between the well (r1 = 0.5 ft) and r2 = 200 ft. Relation: Thiem, for a confined aquifer, Q = 2 pi K b (h2 - h1)/ln(r2/r1). Substitute: T = Kb = 1000 ft^2/day; Q = 2 pi(1000)(4)/ln(200/0.5). Result: Q = 4195 ft^3/day. Check: the well is given by its diameter, so r1 is half of it.
Why A is tempting
Inverted the radius ratio, which flips the sign of the logarithm.
Why B is tempting
Used pi rather than 2pi. Confined radial flow enters the well through the full circumference.
Why D is tempting
Used a base-10 logarithm where the derivation calls for a natural logarithm.
Source FHWA HEC-22, Urban Drainage Design Manual, 3rd ed. — Chapters 3 and 7, hydrology and storm drains
Four mistakes that cost the question
Each of these lands on an answer that is offered, so it costs the question outright. The minutes are our estimate of the time each one burns on top, against an average of 2.9 minutes a question (320 minutes for 110).
Costs the question and about 3 min
Taking the slope from the wrong elevations
For uniform flow, the slope in Manning's equation is the pipe or channel slope: upstream invert minus downstream invert, over the length between them. Using ground or rim elevations gives the wrong slope, and the square root in the equation softens the error just enough that the result is offered.
Costs the question and about 3 min
Assuming a half-full pipe carries half the velocity
A circular pipe flowing half full has a hydraulic radius of D/4, the same as flowing full, so the velocity is the same and the discharge is half because the area is half. Answers that halve the velocity, or keep the full discharge, are both offered.
Costs the question and about 2 min
Using the wrong storm duration in the rational method
Q = CiA gives cfs directly with the intensity in in/hr and the area in acres. The intensity must be for a storm lasting the time of concentration; reading the curve at any other duration gives a peak that is plausible and wrong.
Costs the question and about 2 min
Dropping the 8.34 in a loading calculation
Pounds per day of a chemical or pollutant is flow in MGD times concentration in mg/L times 8.34. A chlorine feed must also cover demand plus the residual that has to be left; dosing only the demand is a ready-made distractor.
Questions about Water Resources and Environmental Engineering
Why does Water Resources and Environmental cover so much ground?
Its outline has eleven sub-topics, from hydrology and hydraulics through pumps, distribution, flood control, stormwater, collection systems, groundwater, water quality, testing and treatment. It also carries one of the largest question ranges on the paper, so breadth matters more than depth here.
Which open-channel equation should I default to?
Manning's. In US units the velocity is 1.49 divided by n, times the hydraulic radius to the two-thirds, times the square root of the slope. The 1.49 converts the SI form; leave it out and the velocity is a third too small. Pressure pipe flow uses Hazen-Williams or Darcy-Weisbach, and the coefficient the question gives tells you which.
How much treatment and water quality is on the exam?
Water quality, testing and standards, and treatment are three of the eleven sub-topics. Questions are mostly mass balance and loading: detention time, surface overflow rate, weir loading, food-to-microorganism ratio, BOD and chemical dose in pounds per day.
What pump questions should I expect?
Total dynamic head from elevations and losses, power from flow, head and efficiency, net positive suction head available from atmospheric, vapour and suction terms, and the affinity laws for a change in speed: flow with speed, head with its square, power with its cube.
How many study hours this area is worth · Using the handbook under the clock