NCEES PE Agricultural and Biological Engineering · USA · NCEES (ASABE is the content partner)
Original practice questions written to the published syllabus · not past papersHow much water a crop needs, when to give it, how to measure and convey it, and how to design surface irrigation.
In the exams: GATE AG Section 5 (irrigation depth, interval and command area in 5 of 8 papers 2019–26); ICAR AIEEA-PG Unit VII; ICAR AICE subject 43; NCEES PE irrigation; PRC ABE; IA CAIS
How our questions for NCEES PE Agricultural and Biological Engineering spread across the units of Irrigation Engineering (Soil-Water-Plant, Conveyance, Surface Methods). Difficulty: 3 easy, 20 medium, 9 hard.
| Underground pipeline systems | 5 | |
| Furrow irrigation | 4 | |
| Irrigation efficiencies and distribution uniformity | 4 | |
| Irrigation scheduling | 3 | |
| Water measurement: weirs, Parshall flume and orifices | 3 | |
| Border irrigation | 3 | |
| Soil-water-plant relations: field capacity, wilting point and MAD | 2 | |
| Evapotranspiration and FAO-56 Penman-Monteith | 2 | |
| Basin irrigation | 2 | |
| Crop coefficients and crop water requirement | 1 | |
| Effective rainfall | 1 | |
| Design of irrigation channels | 1 | |
| Command area development | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
The permanent wilting point is conventionally set at −1500 kPa. Expressed as pressure head, this is closest to:
D. −153 m of water
1 kPa ≈ 0.102 m of water, so −1500 kPa ≈ −153 m. −15.3 m is a factor-of-ten slip (it is −150 kPa); −1500 m treats kPa as metres. −1.53 m is a factor-of-100 slip.
Further reading: Michael, Irrigation: Theory and Practice
Where should tensiometers be installed to schedule a deep-rooted row crop?
B. One in the upper root zone to trigger irrigation and one near its bottom to check that water reached it without excess drainage
Most extraction is in the upper root zone, so the trigger sensor goes there; a deeper sensor shows whether the irrigation refilled the lower zone or drained past it. Surface readings fluctuate with evaporation, and sensors below the roots do not see crop use.
Further reading: Irrigation Association scheduling guidance; Michael, Irrigation
A 3-ft Parshall flume reads Ha = 0.50 ft and Hb = 0.40 ft. Which is correct?
C. Submergence is 80 %, above the 70 % free-flow limit for this size, so a submergence correction must be applied
Submergence = Hb/Ha = 0.40/0.50 = 80 %. For 1–8 ft flumes free flow holds up to 70 %, so the flow is submerged and the free-flow rating overstates discharge unless corrected. 125 % inverts the ratio; 20 % is 1 − Hb/Ha.
Further reading: USBR Water Measurement Manual
Along an undulating buried irrigation pipeline, where should air release valves and drain valves be placed?
D. Air release at high points; drain valves at low points
Air collects at summits and can block flow or cause surges, so air valves go at high points (and vacuum relief to prevent collapse on draining). Water collects at low points, where drains allow emptying for maintenance and freezing protection.
Further reading: USDA NRCS NEH Part 623
A buried plastic pipeline 300 m long, 200 mm internal diameter, carries 30 L/s. Using Hazen-Williams with C = 150 (SI: hf = 10.67 L Q^1.852 / (C^1.852 D^4.87)), what is the friction loss?
C. 1.14 m
hf = 10.67 × 300 × 0.03^1.852 / (150^1.852 × 0.2^4.87) = 1.14 m. 10× results from taking L as 3,000 m; 0.039 m uses a 400 mm pipe (diameter doubled); 2.43 m uses C = 100, a value for old rough pipe, not plastic.
Further reading: Hazen-Williams; USDA NRCS NEH Part 623
Water flowing at 1.5 m/s in a plastic pipeline (pressure wave speed 400 m/s) is stopped instantly by a valve. What is the theoretical pressure rise by the Joukowsky equation?
B. 600 kPa (about 61 m of water)
ΔP = ρ c ΔV = 1,000 × 400 × 1.5 = 600,000 Pa = 600 kPa ≈ 61 m. 60 and 6,000 kPa are factor-of-ten slips; 300 kPa halves the result, which applies to neither instant nor slow closure.
Further reading: Joukowsky; Streeter and Wylie, Fluid Mechanics
In US customary units the empirical maximum non-erosive furrow stream is about 10/S gallons per minute, with S in percent. For a furrow on a 0.5 % slope this is:
C. 20 gpm
10 / 0.5 = 20 gpm. 5 gpm multiplies; 2,000 gpm uses S as a fraction (0.005); 0.05 gpm inverts the relation.
Further reading: USDA NRCS NEH Part 623
Catch-can depths along a field (mm) are 10, 12, 8, 14, 6 and 10. What is Christiansen's coefficient of uniformity?
B. 80.0 %
Mean = 10 mm; absolute deviations 0, 2, 2, 4, 4, 0 sum to 12. CU = 100 (1 − 12 / (6 × 10)) = 80.0 %. 60 % is the minimum divided by the mean; −20 % forgets to divide by n; 86.7 % uses the range instead of summed deviations.
Further reading: Christiansen (1942)
Infiltrated depths (mm) at eight equally spaced points are 22, 30, 26, 34, 18, 28, 32 and 30. What is the low-quarter distribution uniformity?
D. 72.7 %
Mean = 220 / 8 = 27.5 mm. The lowest quarter (2 of 8 values) is 18 and 22, mean 20 mm. DU = 20/27.5 = 72.7 %. 65.5 % uses only the minimum; 85.5 % averages the lowest half; 58.8 % divides by the maximum.
Further reading: Irrigation Association; Burt et al. (1997)
An irrigation must apply at least 80 mm to the driest quarter of a field. The system's low-quarter DU is 0.80. What average depth must be infiltrated?
C. 100 mm
Average = required low-quarter depth / DU = 80 / 0.80 = 100 mm. 64 mm multiplies by DU; 80 mm assumes perfect uniformity; 144 mm adds 80 % of 80 to 80.
Further reading: Irrigation Association
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