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 papersManaging water at the scale of reservoirs, canals and river basins, including floods and farmer participation.
In the exams: ICAR AICE subject 40 Units 7–8; ICAR AIEEA-PG Unit VII; electives in the ICAR 6th Deans' Committee B.Tech
ICAR AIEEA-PG — Agricultural Engineering & Technology, Unit VII
Definitions and short calculations: storage zones, Bligh and Khosla, outlet flexibility, rational method, Gumbel, risk, Muskingum coefficients, project classification and canal operation terms.
How our questions for ICAR AIEEA-PG spread across the units of Water Resources Management, Floods & Canal Systems. Difficulty: 29 easy, 23 medium, 14 hard.
| Diversion works | 12 | |
| Flood estimation | 9 | |
| Reservoirs | 8 | |
| Canal operation | 8 | |
| Water resources of a country | 7 | |
| Water quality management | 7 | |
| Minor irrigation | 6 | |
| Flood control measures | 5 | |
| Water users' associations and participatory irrigation management | 4 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A project costs 500 million currency units, has a 50-year life, annual operation and maintenance of 10 million and annual benefits of 60 million. At 8% interest, using the capital recovery factor, what is the benefit-cost ratio?
C. 1.18
CRF = 0.08 × 1.08⁵⁰ ÷ (1.08⁵⁰ − 1) = 0.0817. Annual capital cost = 500 × 0.0817 = 40.9; total annual cost = 40.9 + 10 = 50.9; BCR = 60 ÷ 50.9 = 1.18. 3.00 ignores interest (500/50 + 10 = 20); 1.47 leaves out O&M (60 ÷ 40.9); 0.12 divides benefits by the full capital cost.
Further reading: Loucks and van Beek, Water Resource Systems Planning and Management
A catchment of 800 km² yields 1,200 t/km² of sediment a year to a reservoir with 95% trap efficiency. The deposit's dry bulk density is 1.2 t/m³. What volume of sediment is deposited each year?
B. 0.76 Mm³
Mass trapped = 800 × 1,200 × 0.95 = 912,000 t; volume = 912,000 ÷ 1.2 = 760,000 m³. 0.80 omits trap efficiency; 0.91 forgets to divide by bulk density; 1.09 multiplies by 1.2 instead of dividing.
Further reading: Morris and Fan, Reservoir Sedimentation Handbook
A reservoir of 50 Mm³ receives 0.4 Mm³ of sediment a year (as deposited volume), with a constant trap efficiency of 90%. How long until half of its capacity is lost?
C. About 69 years
Deposited per year = 0.4 × 0.9 = 0.36 Mm³; half capacity = 25 Mm³; time = 25 ÷ 0.36 = 69.4 years. 63 ignores trap efficiency; 139 is for the full capacity; 56 multiplies by 0.9 instead of dividing. In practice trap efficiency falls as capacity is lost, so life is computed in steps.
Further reading: Morris and Fan, Reservoir Sedimentation Handbook
Why do designers prefer proportional or slightly sub-proportional outlets near the tail of a distributary?
C. So that when the distributary runs low, head outlets do not take more than their share and starve the tail
Fair sharing of fluctuations protects the tail. A hyper-proportional outlet upstream takes extra when levels rise and the effect cascades. No outlet delivers when the channel is closed; downstream control would make the outlet non-modular.
Further reading: Garg, Irrigation Engineering and Hydraulic Structures
Using Ryves' formula Q = C·A^(2/3) with C = 6.8 (Q in m³/s, A in km²), what is the peak flood from a 500 km² catchment?
D. About 428 m³/s
Q = 6.8 × 500^(2/3) = 6.8 × 63.0 = 428 m³/s. 719 uses Dickens' exponent 3/4; 3,400 uses A to the power 1; 152 uses A^0.5.
Further reading: Subramanya, Engineering Hydrology
Which problem is commonly caused by long flood embankments along an aggrading alluvial river?
D. The river bed rises between the embankments, increasing the consequence of a breach
Confined sediment deposits between embankments, raising the bed and flood levels, so breaches become more damaging; embankments also cause drainage congestion behind them. They store little volume, so they do not reduce downstream floods.
Further reading: WMO/GWP Integrated Flood Management
A tubewell yields 30 L/s and runs 10 h a day. Peak crop demand is 6 mm/day and application efficiency is 75%. What area can it serve at peak?
B. 13.5 ha
Daily volume = 0.030 × 36,000 = 1,080 m³; effective = 810 m³; area = 810 ÷ 0.006 = 135,000 m² = 13.5 ha. 18.0 ignores efficiency; 5.4 uses 4 h; 1.35 slips a decimal.
Further reading: Garg, Irrigation Engineering and Hydraulic Structures
What distinguishes irrigation management transfer from participatory irrigation management?
B. Transfer hands management of part or all of a system to users; participation involves them alongside the agency
PIM involves users in planning, operation, maintenance and finance; IMT shifts responsibility (sometimes ownership) to user organisations, usually below a stated level. Neither is defined by fee level or water source.
Further reading: FAO / IWMI reviews of irrigation management transfer
Which Indian state passed the Farmers' Management of Irrigation Systems Act in 1997, an early state law creating water users' associations?
B. Andhra Pradesh
Andhra Pradesh's 1997 Act created WUAs across its irrigation commands and was followed by other states' laws. The other states passed their own laws at different times.
Further reading: Government of Andhra Pradesh, 1997
What is the WHO drinking water guideline value for nitrate?
A. 50 mg/L as nitrate (NO3)
WHO gives 50 mg/L as NO3, roughly 11 mg/L as NO3-N. 10 mg/L is the US limit expressed as nitrate-nitrogen, not as nitrate; 50 mg/L as N would be about 220 mg/L as nitrate; 1.5 mg/L is the fluoride value.
Further reading: WHO Guidelines for Drinking-water Quality
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