NCEES PE Agricultural and Biological Engineering · USA · NCEES (ASABE is the content partner)
Original practice questions written to the published syllabus · not past papersDesigning and evaluating sprinkler and drip systems, from emitter hydraulics to pump selection.
In the exams: GATE AG Section 5 (sprinkler in 4 of 8, drip in 3 of 8 papers 2019–26); ICAR AICE subject 43 Unit 4; NCEES PE irrigation; IA CID (General Ag, Drip/Micro and Sprinkler exams)
NCEES PE Agricultural and Biological Engineering
Design problems: lateral and main sizing with elevation, inlet pressure, TDH and pump power, pump curves and affinity laws, uniformity targets, economic comparison with annual cost.
How our questions for NCEES PE Agricultural and Biological Engineering spread across the units of Sprinkler & Micro-Irrigation Systems. Difficulty: 1 easy, 27 medium, 18 hard.
| System layout and pump selection | 6 | |
| Drip lateral and sub-main design | 5 | |
| Christiansen's uniformity coefficient | 4 | |
| Sprinkler lateral and main design | 4 | |
| Emission uniformity | 4 | |
| Sprinkler spacing and application rate | 3 | |
| Allowable pressure variation and the friction factor F | 3 | |
| Emitter hydraulics: exponent and coefficient of variation | 3 | |
| Fertigation | 3 | |
| Economics of pressurised irrigation | 3 | |
| Sprinkler types and components | 2 | |
| Drip system components | 2 | |
| Filtration | 2 | |
| Automation | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A sprinkler nozzle of 5 mm diameter with discharge coefficient 0.95 operates at 300 kPa. What is its discharge? (g = 9.81 m/s², ρ = 1000 kg/m³)
C. 0.457 L/s
Head h = 300 000/(1000 × 9.81) = 30.58 m. Area a = π/4 × 0.005² = 1.963e-05 m². q = Cd a √(2gh) = 0.95 × 1.963e-05 × √(2 × 9.81 × 30.58) = 4.569e-04 m³/s = 0.457 L/s. Using 300 directly as metres of head gives 1.431 L/s; using the diameter as a radius gives 1.828 L/s; omitting Cd gives 0.481 L/s.
Further reading: Michael, Irrigation: Theory and Practice
Sprinklers each discharge 1.8 m³/h and laterals are 18 m apart. The soil's basic intake rate is 8 mm/h. What is the maximum spacing of sprinklers along the lateral?
C. 12.5 m
From I = 1000 q / (Sl × Sm) with q in m³/h: Sl = 1000 × 1.8 / (8 × 18) = 12.5 m; in practice the next smaller pipe-length multiple (for example 12 m) is chosen. Using 3600 (the L/s constant) with m³/h gives 45 m; converting 1.8 m³/h to 0.5 L/s and then using 1000 gives 3.47 m. 25.0 m uses half the lateral spacing, 1000 × 1.8/(8 × 9).
Further reading: Michael, Irrigation: Theory and Practice
Runoff appears only near the outer end of a centre pivot on a loam, although the depth per revolution matches crop need. Which change addresses the cause most directly?
B. Spread the application over a wider wetted width (booms or wider-throw heads) or add surface storage by tillage
At the outer end the instantaneous application rate is highest because the same wetted width must deliver the largest discharge. Speeding up reduces depth per pass but not the instantaneous rate; reducing outer discharge under-irrigates the largest area. Widening the wetted pattern or storing surface water attacks the rate problem.
Further reading: Keller and Bliesner, Sprinkle and Trickle Irrigation
Twelve catch cans record (mm): 14, 18, 16, 12, 20, 15, 17, 16, 13, 19, 15, 17. What is Christiansen's CU?
C. 88.5 %
Mean = 192/12 = 16 mm. Σ|xi − x̄| = 22 mm. CU = 100 (1 − 22/(12 × 16)) = 88.5 %. 81.25 % is the low-quarter DU; 75 % is the minimum-to-mean ratio; 85.8 % uses the standard deviation instead of the mean absolute deviation.
Further reading: Michael, Irrigation: Theory and Practice
Emitters have Cv = 0.05, there are 2 emitters per plant, and qmin/qavg in the block is 0.93. What is the design emission uniformity (Keller–Karmeli)?
A. 88.8 %
EU = 100 (1 − 1.27 Cv/√n)(qmin/qavg) = 100 (1 − 1.27 × 0.05/1.414) × 0.93 = 100 × 0.9551 × 0.93 = 88.8 %. Omitting √n gives 87.1 %; omitting qmin/qavg gives 95.5 %; omitting 1.27 gives 89.7 %.
Further reading: ASABE EP405; Keller and Karmeli (1974)
A grower chlorinates drip water from a well with pH 8.3 to control bacterial slime, but results are poor. Which change most improves chlorine's effectiveness?
D. Inject acid to lower pH, since hypochlorous acid predominates at lower pH
Chlorine in water is present as hypochlorous acid (a strong biocide) and hypochlorite ion (weak); the balance shifts towards hypochlorous acid as pH falls, so acid improves effectiveness. Mixing chlorine with fertilisers can cause reactions and is avoided.
Further reading: Lamm, Ayars and Nakayama, Microirrigation for Crop Production
5 kg N/ha is to be applied to 4 ha as urea (46 % N) dissolved at 0.3 kg urea per litre of stock solution and injected over 2 hours. What is the injection rate?
B. 72.5 L/h
N = 5 × 4 = 20 kg; urea = 20/0.46 = 43.5 kg; stock volume = 43.5/0.3 = 145 L; injection rate = 145/2 = 72.5 L/h. Forgetting to convert N to urea gives 33.3 L/h; forgetting the injection time gives 145 L/h. 66.7 L/h forgets both the urea conversion and the 2-hour injection time (20/0.3).
Further reading: Lamm, Ayars and Nakayama, Microirrigation for Crop Production
A drip system has static lift 20 m, well drawdown 3 m, mainline friction 4 m, filter loss 5 m (dirty limit), injector loss 2 m, required sub-main inlet pressure 15 m and a 2 m rise to the critical block. What is the total dynamic head?
A. 51 m
TDH = 20 + 3 + 4 + 5 + 2 + 15 + 2 = 51 m. Omitting drawdown gives 48 m, omitting the elevation rise gives 49 m, and omitting the required operating pressure gives 36 m, all of which would leave the critical block under-pressured.
Further reading: Keller and Bliesner, Sprinkle and Trickle Irrigation
How is the operating point of a centrifugal pump in an irrigation system found?
A. At the intersection of the pump head–flow curve and the system head curve
The pump delivers the flow at which the head it produces equals the head the system requires. The designer then checks that this point is close to best efficiency; if it is not, the pump or impeller is changed.
Further reading: Keller and Bliesner, Sprinkle and Trickle Irrigation
A pump installed with a high suction lift makes a crackling noise and its impeller shows pitting. Which condition explains this?
C. NPSH available is too low
Cavitation occurs when NPSH available falls below NPSH required: vapour bubbles form in the eye of the impeller and collapse, causing noise and pitting. Remedies include reducing suction lift or suction losses.
Further reading: Keller and Bliesner, Sprinkle and Trickle Irrigation
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