FE Civil Water Resources and Environmental Engineering practice problems
6 free FE Civil Water Resources and Environmental Engineering practice problems with answers and full worked solutions. Water Resources and Environmental Engineering carries 10–15 of the 110 questions on the FE Civil exam (NCEES specification effective July 2020). Pick an answer to check it, then read the worked solution and why each wrong choice is wrong.
6 worked Water Resources and Environmental Engineering problems
Question 1 of 6
A watershed has an NRCS curve number of 90. For a storm producing 6.5 in of rainfall, the depth of direct runoff is most nearly:
Show the answer and the reasoning
Why B
Given: curve number 90, rainfall P = 6.5 in. Relation: S = 1000/CN - 10 and Ia = 0.2S; once P exceeds Ia, Q = (P - 0.2S)^2/(P + 0.8S). Substitute: S = 1.11 in, Ia = 0.22 in; Q = (6.5 - 0.22)^2/(6.5 + 0.89). Result: Q = 5.33 in.
Why A is tempting
Used S rather than 0.8S in the denominator.
Why C is tempting
Left the initial abstraction out of the numerator.
Why D is tempting
Treated the curve number as a straight runoff percentage.
Source FHWA HEC-22, Urban Drainage Design Manual, 3rd ed. — Section 3.2.3, NRCS runoff
Question 2 of 6
A rectangular channel 6 ft wide carries 320 cfs. The critical depth is most nearly:
Rectangular open channel 6 ft wide carrying 320 cubic feet per second, with the flow depth measured from the invert to the water surface. Show the answer and the reasoning
Why B
Given: Q = 320 cfs in a rectangular channel 6 ft wide. Relation: yc = (q^2/g)^(1/3), with q = Q/b, the discharge per foot of width. Substitute: q = 320/6 = 53.33 cfs/ft; yc = (53.33^2/32.2)^(1/3). Result: yc = 4.45 ft. Check: at this depth the Froude number is exactly 1.
Why A is tempting
Used 2g. The critical-depth relation carries g, not 2g.
Why C is tempting
Confused critical depth with the minimum specific energy, which is 1.5 times the critical depth.
Why D is tempting
Used the total discharge instead of the discharge per unit width.
Source FHWA HDS-4, Introduction to Highway Hydraulics — Section 2.4, critical flow
Question 3 of 6
A pump delivers 22 cfs against a total dynamic head of 50 ft. The pump efficiency is 70%. The brake horsepower required is most nearly:
Show the answer and the reasoning
Why C
Given: Q = 22 cfs, total dynamic head 50 ft, efficiency 0.7. Relation: water horsepower = Q gamma H/550; brake horsepower = water horsepower/efficiency. Substitute: whp = (22)(62.4)(50)/550 = 124.8 hp; bhp = 124.8/0.7. Result: 178 hp.
Why A is tempting
Used 33,000 ft-lb/min. With Q in cfs the power is per second, so the constant is 550 ft-lb/s.
Why B is tempting
Reported the water horsepower. Brake horsepower is the shaft power the motor must deliver, so divide by the efficiency.
Why D is tempting
Divided by 0.746, converting toward kilowatts when horsepower was asked for.
Source FHWA HEC-22, Urban Drainage Design Manual, 3rd ed. — Chapters 3 and 7, hydrology and storm drains
Question 4 of 6
A 16-in diameter water main carries 1350 gpm. The average velocity in the pipe is most nearly:
Water main of 16 inches inside diameter carrying 1,350 gallons per minute. Show the answer and the reasoning
Why D
Given: 1350 gpm in a 16-in main. Relation: V = Q/A, with Q in cfs (448.83 gpm per cfs) and A = pi d^2/4 in ft^2. Substitute: Q = 1350/448.83 = 3.008 cfs; A = pi(16/12)^2/4 = 1.3963 ft^2. Result: V = 2.15 ft/s. Check: distribution mains are normally kept below about 5 ft/s.
Why A is tempting
Used piD^2 rather than piD^2/4 for the flow area.
Why B is tempting
Divided by the circumference of the pipe rather than its cross-sectional area.
Why C is tempting
Used the round 1 cfs = 500 gpm. The exact figure is 448.8 gpm.
Source FHWA HEC-22, Urban Drainage Design Manual, 3rd ed. — Chapters 3 and 7, hydrology and storm drains
Question 5 of 6
A 90-degree V-notch weir is used to gauge a small stream. The head above the notch is 1.1 ft. The discharge is most nearly:
Ninety-degree V-notch weir gauging a small stream, with a head of 1.1 ft measured above the notch vertex. Show the answer and the reasoning
Why C
Given: a 90-degree notch with H = 1.1 ft above the vertex. Relation: Q = 2.5 H^(5/2) for a 90-degree V-notch. Substitute: Q = 2.5(1.1)^2.5. Result: Q = 3.17 cfs. Check: the steep exponent is what gives a V-notch its resolution at low flows.
Why A is tempting
Used the 3/2 exponent of a rectangular weir. A V-notch narrows toward the bottom, so its exponent is 5/2.
Why B is tempting
Squared the head instead of raising it to the 5/2 power.
Why D is tempting
Used the rectangular weir coefficient of 3.33.
Source FHWA HEC-22, Urban Drainage Design Manual, 3rd ed. — Section 7.4, V-notch weirs
Question 6 of 6
A development produces a triangular runoff hydrograph with a peak of 240 cfs and a base time of 90 minutes. The permitted release rate is 60 cfs. The detention storage required is most nearly:
Triangular runoff hydrograph peaking at 240 cubic feet per second over a base time of 90 minutes; the area under it is the inflow volume, and the shaded part above the 60 cfs release is what has to be stored. Show the answer and the reasoning
Why D
Given: a triangular inflow peaking at 240 cfs over a 90-min base; permitted release 60 cfs. Relation: storage is the triangle above the release, 0.5 x base x (Qin - Qout), in acre-feet. Substitute: 0.5(90 x 60 s)(180.0 cfs)/43,560. Result: 11.2 ac-ft.
Why A is tempting
Left the base time in minutes without converting to seconds.
Why B is tempting
Halved the volume twice.
Why C is tempting
Took one third of base times height, the factor for a cone. A triangle is one half.
Source FHWA HEC-22, Urban Drainage Design Manual, 3rd ed. — Section 8.4, storage sizing
Going further
Frequently asked questions
How many Water Resources and Environmental Engineering questions are on the FE Civil exam?
Between 10 and 15 of the 110 questions, according to the NCEES FE Civil specification effective July 2020.
Are these real NCEES exam questions?
No. NCEES does not release its exam questions. These are original FE-style problems written to the NCEES specification and the FE Reference Handbook; every calculated key is checked by two independent solution methods.
Are these Water Resources and Environmental Engineering practice problems free?
Yes. All 6 problems on this page, with answers and full worked solutions, are free and need no account.
Check where you stand, free
The quick diagnostic is eight FE Civil questions in about ten minutes, with no account. The free mock is a full 110-question, 5 hour 20 minute FE Civil paper with a worked solution for every question.