ICAR AIEEA-PG — Agricultural Engineering & Technology (Major Subject Group 10) · India · National Testing Agency for ICAR
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)
How our questions for ICAR AIEEA-PG spread across the units of Sprinkler & Micro-Irrigation Systems. Difficulty: 28 easy, 32 medium, 5 hard.
| Sprinkler types and components | 7 | |
| Sprinkler spacing and application rate | 6 | |
| Christiansen's uniformity coefficient | 6 | |
| Sprinkler lateral and main design | 5 | |
| Allowable pressure variation and the friction factor F | 5 | |
| Drip system components | 5 | |
| Fertigation | 5 | |
| Economics of pressurised irrigation | 5 | |
| Emitter hydraulics: exponent and coefficient of variation | 4 | |
| Drip lateral and sub-main design | 4 | |
| Filtration | 4 | |
| System layout and pump selection | 4 | |
| Automation | 3 | |
| Emission uniformity | 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
Eight catch cans between sprinklers collect 10, 12, 9, 11, 8, 12, 10 and 8 mm. What is Christiansen's uniformity coefficient?
D. 87.5 %
Mean = 80/8 = 10 mm. Absolute deviations 0, 2, 1, 1, 2, 2, 0, 2 sum to 10 mm. CU = 100 (1 − 10/(8 × 10)) = 87.5 %. 80 % is the low-quarter DU; 85.0 % uses the standard deviation (a different coefficient). 12.5 % is the deviation ratio 10/80 itself, not 100 (1 − 10/80).
Further reading: Michael, Irrigation: Theory and Practice
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
As the number of outlets on a lateral becomes very large, the F factor for the Darcy-Weisbach equation (m = 2) tends to:
D. 1/3
F → 1/(m+1) = 1/3 for m = 2. 0.351 is the limit for Hazen-Williams (m = 1.852); 1 applies to a pipe with a single outlet at the end.
Further reading: Keller and Bliesner, Sprinkle and Trickle Irrigation
Why do labyrinth (tortuous-path) emitters allow a larger flow passage than long-path laminar emitters for the same discharge?
D. The zigzag path creates turbulence that dissipates head in a shorter, wider channel
In a tortuous path, repeated changes of direction dissipate energy through turbulence, so a wider, shorter channel achieves the same head loss; this reduces clogging risk and makes x close to 0.5. A long laminar path needs a very narrow, long channel.
Further reading: Lamm, Ayars and Nakayama, Microirrigation for Crop Production
In an older drip system that was designed with a high emission uniformity, what is usually the main reason field EU has fallen?
C. Emitter clogging
Physical, chemical and biological clogging reduces flow unevenly and is the dominant cause of declining uniformity. Water treatment, filtration and flushing are the remedies.
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
Which pair of fertilisers must not be mixed in the same concentrated stock tank because a precipitate forms?
C. Calcium nitrate and a sulphate or phosphate fertiliser
Calcium reacts with sulphate to form gypsum and with phosphate to form calcium phosphates, both of low solubility; they are kept in separate A and B tanks. The other pairs are compatible in normal stock concentrations.
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
When dividing a drip field into irrigation shifts (zones), what is the main hydraulic aim?
B. Shifts of about equal flow so the pump always runs near its design point
Equal-flow shifts keep the pump at the same operating point and efficiency for every shift. Equal area is not enough when crops or emitter densities differ.
Further reading: Keller and Bliesner, Sprinkle and Trickle Irrigation
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