NCEES PE Agricultural and Biological Engineering · USA · NCEES (ASABE is the content partner)
Original practice questions written to the published syllabus · not past papersNCEES PE Agricultural and Biological Engineering is a licensing and registration exam run by NCEES (ASABE is the content partner), for candidates in USA. Computer-based, 85 questions, 9.5-hour appointment (8.5 hours of exam). Once a year (next: 28 Oct 2026). Closed book with the electronic NCEES handbook and supplied searchable ASABE standards; SI and US customary units.
Source: ncees.org.
Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.
Managing water at the scale of reservoirs, canals and river basins, including floods and farmer participation.
Our 22 NCEES PE Agricultural and Biological Engineering questions in this subject: 6 easy, 7 medium, 9 hard, across 4 units.
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
These 10 were written to NCEES PE Agricultural and Biological Engineering and are not shown on any other exam's page.
A reservoir with 60 Mm³ of live storage receives monthly inflows of 20, 35, 60, 80, 45, 25, 12, 8, 5, 4, 6 and 10 Mm³ in a repeating year. What is the largest uniform monthly draft it can sustain?
B. 17.5 Mm³ per month
With draft D above 12, the six dry months give deficits summing to 6D − (12 + 8 + 5 + 4 + 6 + 10) = 6D − 45. Setting 6D − 45 = 60 gives D = 17.5 Mm³. 25.8 is the mean inflow, which would need far more storage; 20 needs 75 Mm³.
A reservoir's water spread is 15 km². Class A pan evaporation averages 6 mm/day and the pan coefficient is 0.7. What is the evaporation loss in a 30-day month?
A. 1.89 Mm³
Lake evaporation = 0.7 × 6 = 4.2 mm/day; loss = 15 × 10⁶ m² × 0.0042 m × 30 = 1.89 × 10⁶ m³. 2.70 omits the pan coefficient; 0.063 is one day only; 18.9 slips a decimal.
Logarithms (base 10) of annual peak flows have mean 2.8 and standard deviation 0.25. For the station skew and a 100-year return period, the log-Pearson Type III frequency factor is 2.4. What is the 100-year flood?
A. About 2,510 m³/s
log Q = 2.8 + 2.4 × 0.25 = 3.4; Q = 10^3.4 = 2,512 m³/s. 635 back-transforms the mean and standard deviation separately (10^2.8 + 2.4 × 10^0.25); 1,260 is 10^3.1 (K applied as 1.2); 3,980 is 10^3.6 (K × σ taken as 0.8).
For which structure is the probable maximum flood normally used as the design flood?
C. The spillway of a large dam whose failure would endanger life
PMF, from probable maximum precipitation, is used where failure would be catastrophic, such as large dam spillways. Culverts, waterways and storm drains are designed for return periods of a few to a few tens of years.
A triangular flood hydrograph has a peak of 900 m³/s and a base of 48 h. The channel downstream can safely carry 500 m³/s. What detention storage is needed to keep the flow at or below the safe capacity?
A. About 15.4 Mm³
The part above 500 is a triangle 400 m³/s high and 48 × 400/900 = 21.3 h wide: volume = 0.5 × 21.3 × 3,600 × 400 = 15.4 Mm³. 77.8 is the whole flood volume; 34.6 uses the full 48 h base for the 400 m³/s excess; 7.7 halves the correct volume a second time.
What does the biochemical oxygen demand (BOD) of a water sample measure?
A. The oxygen used by microbes to oxidise organic matter, conventionally over 5 days at 20 °C
BOD is microbial oxygen demand, conventionally BOD5 at 20 °C. Chemical oxidation with a strong oxidant is COD; the dissolved oxygen at collection is DO.
Wastewater has an ultimate BOD of 200 mg/L and a base-e rate constant k = 0.23 per day. Using BOD_t = L0(1 − e^(−kt)), what is BOD5?
B. About 137 mg/L
BOD5 = 200 × (1 − e^(−1.15)) = 200 × 0.683 = 137 mg/L. 186 uses base 10 with the base-e constant (the common mix-up); 46 is 200 × 0.23; 63 is the BOD remaining, not exerted.
A river carrying 20 m³/s at 2 mg/L nitrate receives a drain of 5 m³/s at 22 mg/L. After complete mixing, what is the nitrate concentration?
D. 6 mg/L
C = (20 × 2 + 5 × 22) ÷ (20 + 5) = 150 ÷ 25 = 6 mg/L. 12 is the simple average; 24 adds concentrations; 4.4 divides only the drain's load by the total flow, leaving out the river's own load.
Which agricultural pollutant reaches surface water mainly attached to eroded soil particles?
B. Phosphorus
Phosphorus binds to soil and moves with sediment, so erosion control reduces it. Nitrate is mobile in solution and leaches to groundwater; chloride and sodium are dissolved salts.
In US water quality regulation, what is a TMDL?
A. The largest pollutant load a water body can take and still meet its standard, allocated among sources
A total maximum daily load sets the pollutant budget for an impaired water and allocates it among point and non-point sources under the Clean Water Act. It is not a withdrawal permit, a minimum flow or a tax.
This plan follows where our questions actually fall, so the time goes where the marks are likely to be. It is our reading of the syllabus, not the exam body's weighting.
Read the manual chapters for these first and do every worked question you can find on them.
Aim for one sitting per unit, then mixed questions across them.
Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.
Do these early in revision so they are fast on the day.
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