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.
Removing excess water and salt: surface and subsurface drain design, and reclaiming saline and sodic land.
Our 20 NCEES PE Agricultural and Biological Engineering questions in this subject: 0 easy, 15 medium, 5 hard, across 9 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 subsurface drainage system in a humid region is designed with a drainage coefficient of 3/8 inch per day on 80 acres. What is the design outlet flow?
D. 1.26 ft3/s
One acre-inch = 3,630 ft3. Daily volume = 0.375 x 80 x 3,630 = 108,900 ft3; divide by 86,400 s: 1.26 ft3/s. The shortcut Q = DC x A / 24 gives 1.25, the same within rounding. 30 forgets to divide by 24 hours; the others are divisor slips.
A trapezoidal field collector has bed width 1.0 m, flow depth 0.8 m, side slopes 1.5H:1V, bed slope 0.0004 and Manning's n = 0.03. What is its capacity?
B. 0.69 m3/s
A = by + zy^2 = 0.8 + 1.5 x 0.64 = 1.76 m2. P = b + 2y sqrt(1 + z^2) = 1 + 1.6 x 1.803 = 3.884 m. R = 0.453 m. Q = (1/0.03) x 1.76 x 0.453^(2/3) x 0.02 = 0.69 m3/s. 0.90 ignores the sloping sides in P; 1.17 omits R^(2/3); 6.92 uses S = 0.04.
For flat humid catchments the Cypress Creek formula has the form Q = C A^(5/6). If the drainage area served by a ditch doubles, the design discharge is multiplied by about:
A. 1.78
Q2/Q1 = 2^(5/6) = 1.78. The exponent below 1 reflects the slower, attenuated response of flat land. 2.00 assumes linearity (as a drainage coefficient does); 1.12 uses the exponent 1/6. 1.41 uses a square-root exponent, 2^0.5.
Donnan's equation: K = 0.6 m/day, drain water level 2.0 m above the barrier, midpoint water table 2.8 m above the barrier, drainage rate 5 mm/day. What is the spacing?
B. 42.9 m
L^2 = 4K(b^2 - a^2)/q = 4 x 0.6 x (7.84 - 4.00) / 0.005 = 1843 m2; L = 42.9 m. Using (b - a)^2 = 0.64 drops the flow below drain level; adding squares is wrong; 1843 is L^2.
A granular envelope is designed with the retention rule D15(envelope) / D85(soil) <= 5. If the soil D85 is 0.12 mm, what is the largest allowable envelope D15?
D. 0.60 mm
D15 envelope <= 5 x D85 soil = 5 x 0.12 = 0.60 mm. Dividing gives 0.024 mm, which would be far too fine to be permeable. A second rule (envelope D15 several times the soil D15) keeps the envelope more permeable than the soil. 6.0 mm misreads the soil D85 as 1.2 mm (5 x 1.2).
A collector pipe drains 10 ha at a drainage coefficient of 10 mm/day. Grade 0.2 %, Manning's n = 0.015, flowing full. Using Q = (1/n)(pi D^2/4)(D/4)^(2/3) S^(1/2), the minimum inside diameter is closest to:
D. 193 mm
Q = 0.010 m/day x 100,000 m2 = 1,000 m3/day = 0.01157 m3/s. Solving the full-pipe Manning equation for D with n = 0.015 and S = 0.002 gives D = 193 mm; choose the next commercial size up. 81 mm uses S = 0.2 (per cent read as a fraction); 458 mm uses a discharge ten times too large (cm/day slip); 97 mm confuses radius with diameter.
A contractor installs pipe drains while the soil is very wet. What performance problem is most likely?
B. Smearing and compaction of the trench wall, sealing the soil by the drain
Wet soil smears under the trencher or plough, forming a low-permeability skin that raises entrance resistance. It does not change spacing or cause ochre directly.
An interceptor drain is most effective when:
C. It is deep enough to reach or cut into the impermeable layer
Reaching the barrier intercepts the whole flow; a shallower drain lets part of the flow pass underneath and its effect fades downslope.
Seepage moves down a hillside through a saturated layer 2.5 m thick with K = 1.5 m/day and hydraulic gradient 0.04. What inflow should a 400 m interceptor drain be designed for?
A. 60 m3/day
Per metre of drain: q = K i b = 1.5 x 0.04 x 2.5 = 0.15 m3/day. Over 400 m: 60 m3/day. 0.15 is per metre only; 24 omits the flow thickness; 600 is a factor-of-ten slip.
Relief wells at the landward toe of a levee or dam are installed mainly to:
B. Reduce uplift pressure from underseepage
Relief wells let pressurised underseepage escape in a controlled way, preventing boils and uplift failure.
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.
Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.
Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.
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