Nepal Engineering Council · Civil Engineering · Chapter 7
Irrigation and Drainage
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183 questions in 6 syllabus topics.
7.1 Water demand estimation
32 questions · ACiE0701
1. Duty of water for a crop is defined as the
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Duty is expressed in ha/cumec.
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Answer: B. area irrigated per unit discharge flowing during the base period
Duty is the area (ha) that can be irrigated by one cumec flowing continuously over the base period of the crop. Depth of water over the base period is delta.
2. A wheat crop has a base period of 120 days and a duty of 1800 ha/cumec at the field. The delta of the crop is
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Use Δ = 8.64 B / D.
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Answer: D. 0.576 m
Δ = 8.64 B / D = 8.64 × 120 / 1800 = 0.576 m (D in ha/cumec, B in days).
3. A crop requires a total delta of 0.9 m over a base period of 140 days. Its duty is
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Rearrange Δ = 8.64 B / D.
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Answer: A. 1344 ha/cumec
D = 8.64 B / Δ = 8.64 × 140 / 0.9 = 1344 ha/cumec.
4. For the same crop and the same canal system, the duty of water is
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Compare the discharge being divided into the area at each point.
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Answer: B. greater at the field than at the canal head
Water is lost in conveyance, so the discharge reaching the field is less than that at the head; the same area per unit discharge is larger when the discharge is smaller. Duty at field > duty at outlet > duty at head.
5. The base period of a crop is the
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It is not necessarily the same as the crop period.
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Answer: A. time between the first watering and the last watering before harvest
Base period (B) is the time from the first irrigation to the last irrigation of the crop; it is the B used in the duty-delta relation.
6. Culturable command area (CCA) is the
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Subtract what cannot be cultivated.
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Answer: A. gross command area minus the unculturable area such as roads, ponds and settlements
GCA is the total area that can be commanded; deducting unculturable land gives CCA, which is the area on which crops can be grown.
7. In a project with CCA = 2000 ha, 800 ha is irrigated in the kharif season and 1100 ha in the rabi season. The intensity of irrigation is
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Add the seasonal areas first.
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Answer: D. 95%
Intensity = (area irrigated annually / CCA) × 100 = (800+1100)/2000 × 100 = 95%.
8. An annual intensity of irrigation of 150% on a canal system means that
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Think of sum of seasonal areas divided by CCA.
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Answer: D. each hectare of CCA is irrigated for 1.5 crops a year on average
Intensity above 100% shows multiple cropping: the sum of the seasonal irrigated areas exceeds the CCA.
9. A 75% dependable flow of a river means a flow that
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Read it from a flow duration curve.
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Answer: C. is equalled or exceeded in 75% of the time in the record
Dependability is the percentage of time (or years) a given flow is equalled or exceeded in the flow-duration record; 75% is commonly adopted for irrigation planning.
10. The consumptive use of a crop is 450 mm and the effective rainfall during its growth is 120 mm. If the water application efficiency is 60%, the field irrigation requirement is
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Net requirement first, then divide by application efficiency.
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Answer: A. 550 mm
NIR = CU − Pe = 450 − 120 = 330 mm; FIR = NIR / Ea = 330/0.6 = 550 mm.
11. A crop has a net irrigation requirement of 400 mm. With field application efficiency 80% and conveyance efficiency 75%, the gross irrigation requirement at the canal head is
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Both efficiencies act in series.
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Answer: C. 667 mm
GIR = NIR / (Ea × Ec) = 400/(0.8×0.75) = 667 mm.
12. A canal diverts 5.0 m³/s at its head and 4.1 m³/s is delivered to the outlets. The conveyance efficiency is
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Output over input.
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Answer: D. 82%
Ec = water delivered to the field channels / water entering the canal at head = 4.1/5.0 = 0.82 = 82%.
13. Water delivered to a field is 90 mm and 63 mm is stored in the root zone. The water application efficiency is
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Stored over delivered.
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Answer: A. 70%
Ea = water stored in the root zone / water delivered to the field = 63/90 = 70%.
14. Water storage efficiency of an irrigation is defined as the ratio of water
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The denominator is what the soil could take (the need).
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Answer: B. water stored in the root zone to water needed to fill it to field capacity
Storage efficiency measures how completely the root zone has been refilled; application efficiency instead compares storage with delivery.
15. Water was applied to a field with an average depth of 50 mm and an average numerical deviation from this mean of 5 mm. The water distribution efficiency, Ed = (1 − y/d) × 100, is
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Deviation divided by mean depth, subtracted from one.
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Answer: C. 90%
Ed = (1 − 5/50) × 100 = 90%.
16. Effective rainfall for irrigation planning is the
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Subtract losses from total rainfall.
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Answer: B. part of the rainfall that is stored in the root zone and available for crop use
Runoff and deep percolation are lost to the crop; the remainder retained in the root zone is effective.
17. Using the USDA Soil Conservation Service formula Pe = P(125 − 0.2P)/125 (for P < 250 mm), the effective rainfall for a monthly rainfall P = 100 mm is
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Put P = 100 straight into the formula.
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Answer: B. 84 mm
Pe = 100 × (125 − 20)/125 = 84 mm.
18. Consumptive use of a crop is the
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It is a crop-and-climate quantity, not a conveyance loss.
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Answer: D. evaporation from the soil plus transpiration by the plant
Consumptive use (evapotranspiration) is the water actually used by the crop; the amount held in plant tissue is negligible.
19. The reference evapotranspiration is 5.0 mm/day and the crop coefficient at mid-season is 1.1. The crop evapotranspiration is
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Multiply by the crop coefficient.
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Answer: B. 5.5 mm/day
ETc = Kc × ET0 = 1.1 × 5.0 = 5.5 mm/day.
20. By the Blaney–Criddle formula u = k·p(0.46T + 8.13), with k = 0.85, p = 0.27 and mean temperature T = 27 °C, the consumptive use is nearly
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Evaluate the bracket first.
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Answer: D. 4.7 mm/day
u = 0.85 × 0.27 × (0.46×27 + 8.13) = 0.85 × 0.27 × 20.55 = 4.72 mm/day.
21. Field capacity of a soil is the moisture content
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Think of the state two or three days after heavy irrigation.
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Answer: D. held in the soil after the excess gravitational water has drained away
After free drainage (about 2–3 days) the soil holds capillary water at field capacity.
22. The permanent wilting point is the soil moisture content at which
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It is the lower limit of available moisture.
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Answer: B. plants can no longer extract water and wilt permanently
Below the permanent wilting point (about 15 atm tension) soil moisture is held too tightly for roots; available moisture is the difference between field capacity and this point.
23. A soil has FC = 28%, PWP = 12%, dry density/density of water = 1.5 and root depth 0.9 m. If management allows 50% depletion of the available moisture, the depth of water to be applied per irrigation is
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Don't forget the density ratio and the allowable depletion fraction.
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Answer: C. 108 mm
d = (FC − PWP)/100 × (γd/γw) × D × 0.5 = 0.16 × 1.5 × 900 × 0.5 = 108 mm.
24. If the depth of water applied per irrigation is 108 mm and the peak consumptive use is 6 mm/day, the maximum interval between irrigations is
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Divide depth by daily use.
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Answer: C. 18 days
Frequency = d / Cu = 108 / 6 = 18 days.
25. Compared with a clay loam, a sandy soil needs
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Consider the water-holding capacity.
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Answer: A. smaller depths of water at more frequent intervals
Sand has a low available moisture capacity and a shallow readily available store, so each irrigation must be light and the interval short.
26. A command of 1000 ha has a peak gross water demand of 7 mm/day at the outlets. The continuous discharge required is nearly
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Convert hectares and mm into m³ per day, then to m³/s.
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Answer: C. 0.81 m³/s
Volume/day = 10⁷ m² × 0.007 m = 70 000 m³; Q = 70 000/86 400 = 0.81 m³/s.
27. The discharge needed at the head of a distributary's outlets is 1.8 m³/s. If the conveyance efficiency of the distributary is 75%, its design discharge at the head should be
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Divide by the efficiency.
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Answer: C. 2.4 m³/s
Q = 1.8 / 0.75 = 2.4 m³/s; losses must be added, not subtracted.
28. The capacity factor of a canal is the ratio of
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Mean over maximum.
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Answer: A. mean discharge to the full-supply (maximum) discharge of the canal
Capacity factor = mean supply / canal capacity; time factor = number of days the canal actually flows / total number of days in the period.
29. Net irrigation requirement (NIR) of a crop is
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Subtract free natural supplies from the crop need.
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Answer: C. crop water requirement minus effective rainfall and soil/groundwater contributions
NIR is the water that irrigation must supply at the root zone; adding losses gives the field and gross irrigation requirements.
30. Which loss is classified as a field (application) loss rather than a conveyance loss?
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It happens after water reaches the farm.
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Answer: B. deep percolation below the root zone
Deep percolation and surface runoff occur on the field during application; seepage, evaporation and leakage occur in the conveyance system.
31. A tank can supply 24 million m³ of water at the field level. For a crop of delta 0.8 m, the area it can irrigate is
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Area = volume / depth; convert m² to ha.
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Answer: D. 3000 ha
Area = volume / delta = 24×10⁶ / 0.8 = 30×10⁶ m² = 3000 ha.
32. For a fixed base period, if the duty of a crop at the field increases, its delta
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Use the duty-delta relation.
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Answer: A. decreases
Δ = 8.64 B / D, so for given B the delta is inversely proportional to duty.
7.2 Design of canals
30 questions · ACiE0702
33. In a canal network, the canal that takes off directly from the headworks and carries the whole supply to the branches is the
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It is the top of the hierarchy.
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Answer: D. main canal
The main canal feeds the branch canals and large distributaries; it is a carrier and normally does not irrigate directly.
34. The correct descending order of canals in a network, starting from the headworks, is
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Minors are the smallest canals managed by the department.
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Answer: B. main canal, branch canal, distributary, minor, watercourse
Supply passes from the main canal to branches, then distributaries and minors, and finally through outlets to watercourses (field channels).
35. An outlet in a canal system is the structure that
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It is the last structure managed by the canal authority.
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Answer: B. delivers water from a distributary or minor into a watercourse
An outlet (or 'mogha') connects a distributary/minor to a watercourse and is the point of transfer between the departmental and farmer-managed system.
36. A perennial canal differs from an inundation canal in that a perennial canal
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The presence of a headwork is the key difference.
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Answer: B. is fed from a headwork and supplies water throughout the year
Perennial canals have a headwork (weir/barrage or reservoir) to guarantee a supply; inundation canals draw only during high river stages.
37. The best alignment for the main canal of an irrigation project is generally along the
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It must command land on both its sides.
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Answer: A. watershed (ridge) line
A ridge canal commands the area on both sides by gravity and avoids cross-drainage works because it does not cross natural drains.
38. A contour canal is aligned
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Think of a canal built on the hillside.
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Answer: A. runs roughly along the contours and commands land on one side only
Contour canals run along the contour with a small bed slope; since ground falls on one side only, one side is commanded. They are common in hilly regions.
39. A side-slope canal is one that
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It does not cross drains.
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Answer: A. runs roughly at right angles to the contours and needs no cross-drainage works
A side-slope canal follows the line of steepest ground fall (between a ridge and a drain), so it does not cross natural drains; the bed slope is kept low by falls.
40. Balancing depth of a canal cross-section is the depth of cutting for which
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Cut and fill are in balance.
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Answer: B. cut and fill quantities are exactly equal
At balancing depth, quantities of cut and fill are equal, so no earth need be borrowed or spoiled and the cost is least.
41. A berm in an irrigation canal section is the
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It is a flat bench in the section.
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Answer: C. level strip left between the toe of the bank and the cutting
Berms add stability to banks, provide space for the earth to fall, and reduce seepage by lengthening the percolation path.
42. Kennedy's critical velocity is V₀ = 0.55 m D^0.64. For a depth D = 2.0 m and critical velocity ratio m = 1.0, V₀ is nearly
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Use a calculator for the power 0.64.
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Answer: C. 0.86 m/s
V₀ = 0.55 × 1.0 × 2.0^0.64 = 0.55 × 1.558 = 0.86 m/s.
43. In Kennedy's theory, a canal reach with critical velocity ratio m greater than 1.0 would be expected to
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Compare actual and critical velocity.
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Answer: C. scour
m = V/V₀. If the actual velocity exceeds the critical velocity, the channel scours; below 1.0 it silts.
44. Kennedy's theory assumes that silt is kept in suspension by
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Compare with the eddies in Lacey's theory.
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Answer: A. the vertical component of the eddies generated at the bed
Kennedy considered only the vertical component of bed eddies; Lacey later introduced the effect of the side eddies as well.
45. Lacey's silt factor for a bed material of mean particle size d (in mm) is f =
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It rises with the square root of grain size.
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Answer: C. 1.76 √d
Lacey's silt factor is given by f = 1.76√d, where d is the mean size in mm.
46. Lacey's silt factor for sand of average size 0.25 mm is
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Take the square root of the size in mm.
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Answer: B. 0.88
f = 1.76 √0.25 = 1.76 × 0.5 = 0.88.
47. A canal carries 30 m³/s in alluvium with Lacey's silt factor f = 1.0. The Lacey regime velocity V = (Qf²/140)^(1/6) is nearly
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Take the sixth root.
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Answer: D. 0.77 m/s
V = (30 × 1/140)^(1/6) = 0.2143^(1/6) = 0.77 m/s.
48. For Q = 30 m³/s, the Lacey regime wetted perimeter P = 4.75√Q is nearly
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Wetted perimeter varies as √Q.
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Answer: B. 26.0 m
P = 4.75 × √30 = 4.75 × 5.477 = 26.0 m.
49. For Q = 30 m³/s, V = 0.77 m/s and f = 1.0, Lacey's regime hydraulic mean radius R = 5V²/(2f) is nearly
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Square the velocity first.
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Answer: A. 1.50 m
R = 5 × 0.774² / 2 = 1.50 m.
50. For Q = 30 m³/s and f = 1.0, Lacey's regime bed slope S = f^(5/3)/(3340 Q^(1/6)) is nearly 1 in
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Evaluate Q^(1/6) first.
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Answer: D. 5888
S = 1/(3340 × 30^(1/6)) = 1/(3340 × 1.763) = 1/5888.
51. If the discharge in a Lacey regime channel is increased four times (silt factor unchanged), the regime wetted perimeter becomes
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P varies with the square root of Q.
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Answer: C. two times
P = 4.75√Q, so P ∝ √Q; quadrupling Q doubles P.
52. A major difference between Kennedy's and Lacey's methods is that Lacey's method
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Think about the design procedure.
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Answer: D. gives direct formulae for V, A and S using a silt factor, with no trial and error
Lacey's regime equations relate V, P, R and S directly with Q and f, while Kennedy's design needs trial and error with m and the Kutter formula.
53. In Lacey's theory, a channel is in 'true regime' when
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Consider the constants required.
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Answer: A. Q and silt charge are constant in incoherent alluvium of the same grade
Lacey's regime conditions require constant discharge, constant silt charge and grade, and unlimited incoherent alluvium; real channels only approximate these.
54. The average boundary shear (tractive force) in a canal, τ₀ = γRS. For R = 1.5 m and bed slope 1 in 5000, τ₀ is nearly
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Use γ = 9810 N/m³.
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Answer: D. 2.94 N/m²
τ₀ = 9810 × 1.5 × 0.0002 = 2.94 N/m².
55. In the tractive force approach, a channel in non-cohesive soil is stable if
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Compare applied and permissible shear.
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Answer: C. the tractive force of flow is within the critical tractive force of the material
Bed and side particles move only if the shear stress exceeds the permissible (critical) tractive force for that material and slope.
56. For a trapezoidal canal section, the maximum tractive force occurs on the bed and is nearly 0.97γDS, while on the side slopes it is nearly
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Side shear is less than bed shear.
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Answer: B. 0.75γDS
For trapezoidal sections the maximum shear on the sides is about 75% of γDS, so the sides are less liable to scour than the bed.
57. Side slopes of 2H:1V (θ = 26.57°) are used in a canal in non-cohesive soil of angle of repose φ = 35°. The ratio of critical tractive force on the side to that on the bed, K = √(1 − sin²θ/sin²φ), is nearly
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Evaluate both sines and square them.
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Answer: C. 0.63
sin θ = 0.447, sin φ = 0.574; K = √(1 − 0.608) = 0.63.
58. A trapezoidal canal has bed width 6 m, full-supply depth 1.5 m, side slopes 1.5H:1V, n = 0.025 and bed slope 1 in 4000. By Manning's formula the discharge is nearly
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Compute A, P and R first.
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Answer: D. 8.3 m³/s
A = (6 + 1.5×1.5)×1.5 = 12.375 m²; P = 6 + 2×1.5×√3.25 = 11.41 m; R = 1.085 m; Q = (1/0.025)×12.375×1.085^(2/3)×(1/4000)^0.5 = 8.3 m³/s.
59. The most efficient (economical) trapezoidal section for a concrete-lined canal of depth 2.0 m and side slopes 1H:1V has a bed width b = 2y(√(1+z²) − z) of nearly
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Use b = 2y(√(1+z²) − z).
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Answer: A. 1.66 m
b = 2×2×(√2 − 1) = 4 × 0.4142 = 1.66 m.
60. For the most economical trapezoidal section the hydraulic mean radius is equal to
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R = y/2.
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Answer: D. half the depth of flow
For all best hydraulic trapezoidal sections R = y/2; the half-hexagon (z = 1/√3) is the best of all.
61. Which of the following is NOT an advantage of lining an irrigation canal?
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One of these is a disadvantage.
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Answer: A. lower initial cost of construction
Lining reduces seepage and waterlogging and allows higher velocities, but its initial cost is higher than that of an unlined canal.
62. The value of Manning's n generally adopted for a well-finished concrete-lined canal is about
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Smooth lining has a low roughness.
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Answer: B. 0.015
Smooth concrete has n of about 0.014–0.016; earth canals are usually taken as 0.0225–0.025.
7.3 Diversion headworks
32 questions · ACiE0703
63. The main difference between a weir and a barrage is that
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Consider how the pond level is controlled.
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Answer: D. a barrage regulates the pond level by gates, a weir by its fixed crest
A weir raises the water by its crest, with a fixed pond level; a barrage holds most of the head by gates, so afflux during floods is small.
64. The purpose of the divide wall in a diversion headworks is to
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It divides two parts of the structure.
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Answer: C. separate the undersluices from the weir and keep turbulence out of the pocket
The divide wall, perpendicular to the weir axis, separates the scouring sluices (undersluices) from the weir and gives a quiet pocket in front of the canal head regulator.
65. Undersluices (scouring sluices) in a diversion headwork are provided mainly to
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They are kept at a low level on the river side of the head regulator.
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Answer: D. flush the silt deposited in the pocket and pass low floods
Undersluices are low-level gated openings next to the canal head; opening them keeps a clear channel to the canal and flushes silt.
66. The crest of the canal head regulator is kept higher than the crest of the undersluices so that
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Sediment moves along the bottom.
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Answer: B. bed load is passed by the undersluices and clearer water enters the canal
Sediment travels near the bed; a raised canal sill (about 1–1.5 m above the undersluice crest) keeps the coarse bed load out of the canal.
67. A fish ladder in a diversion headworks is provided to
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Think about aquatic life.
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Answer: A. allow fish to migrate upstream past the weir or barrage
A fish ladder is a series of stepped pools or channels with low velocities that fish can ascend; it is placed beside the divide wall or at the river bank.
68. Marginal (afflux) bunds on both banks upstream of a weir are provided to
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They protect against the rise in flood level.
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Answer: D. protect land from the afflux and prevent outflanking
Marginal bunds connect the weir to high ground upstream, confining the raised flood level and preventing flow around the structure.
69. Afflux of a weir is the
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It concerns the flood level, before and after construction.
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Answer: C. rise in the highest flood level upstream caused by constructing the weir
Afflux is HFL after construction minus HFL before; it is reduced in barrages by opening the gates.
70. A silt excluder differs from a silt ejector in that the excluder
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One is upstream of the regulator, the other downstream.
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Answer: B. removes sediment before water enters the head regulator, the ejector after it enters the canal
Excluders (tunnels in the pocket upstream of the head regulator) divert bed load to the undersluices; ejectors (extractors) are built in the canal downstream of the head to extract sediment already in it.
71. The basic principle of a settling basin is to
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Bigger section, slower flow.
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Answer: C. reduce the flow velocity so that suspended particles settle out
In a settling basin the cross-section is enlarged, velocity reduced, and sediment settles at its fall velocity during the flow-through time.
72. For an ideal settling basin handling Q = 10 m³/s, to trap particles of fall velocity 0.02 m/s the minimum plan area is
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Use A = Q / ω.
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Answer: B. 500 m²
For a particle to settle, surface area ≥ Q/ω = 10 / 0.02 = 500 m², irrespective of depth.
73. A settling basin has flow depth 4 m, mean flow velocity 0.3 m/s and the design particle has a fall velocity 0.03 m/s. The minimum length of the basin is
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A particle must reach the floor before the basin ends.
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Answer: B. 40 m
Time to settle = h/ω = 4/0.03 = 133 s; length = V × t = 0.3 × 133 = 40 m (L = hV/ω).
74. According to Bligh's creep theory, the percolating water follows the outline of the base of the structure and
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Bligh did not differentiate between the directions of creep.
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Answer: A. head loss is proportional to creep length, horizontal and vertical creep weighted equally
Bligh assumed a uniform hydraulic gradient along the total creep length (H/L = 1/C) without distinguishing between vertical and horizontal paths.
75. By Bligh's theory, the minimum creep length required for a weir with a differential head of 4 m on coarse sand (C = 12) is
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Creep length = C × H.
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Answer: D. 48 m
L = C H = 12 × 4 = 48 m.
76. Bligh's creep coefficient C is largest for
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Finer soil is more prone to piping.
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Answer: A. very fine sand and silt
Bligh's C ranges from about 18 for very fine sand and silt to 4–9 for gravel and boulders; finer soils need longer creep path.
77. By Bligh's theory, an impervious floor at a point where the residual uplift head is 2.4 m, having a specific gravity of 2.4, needs a thickness t = (4/3) h/(G − 1) of nearly
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Net weight of floor must balance uplift.
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Answer: B. 2.29 m
t = 1.33 × 2.4 / (2.4 − 1) = 2.29 m (the 4/3 factor allows 33% for safety).
78. In Lane's weighted creep theory, the weighted creep length is the sum of
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Horizontal paths are less effective.
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Answer: B. vertical creep (steeper than 45°) plus one-third of horizontal creep
Lane found that water creeps more easily along horizontal contacts, so Lw = Lv + Lh/3.
79. A weir floor has total vertical creep 9 m (including steep cut-offs) and horizontal creep 45 m. Lane's weighted creep length is
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Divide the horizontal part by 3.
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Answer: C. 24 m
Lw = Lv + Lh/3 = 9 + 45/3 = 24 m.
80. For a head of 5 m on fine sand (Lane's weighted creep ratio Cw = 7), the minimum weighted creep length required is
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Multiply the creep ratio by the head.
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Answer: D. 35 m
Lw ≥ Cw × H = 7 × 5 = 35 m.
81. Khosla's method of independent variables is based on
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It treats seepage as potential flow.
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Answer: C. the analytical solution of the Laplace equation for seepage flow beneath the floor
Khosla solved potential flow for standard profiles (pile, depressed floor) by conformal mapping and applied corrections for actual shapes; it gives uplift and exit gradient.
82. Khosla gives the exit gradient for a floor of length b with a downstream cut-off of depth d under head H as GE = (H/d)·1/(π√λ), λ = [1 + √(1+α²)]/2, α = b/d. For H = 4 m, d = 6 m, b = 30 m, GE is nearly
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Compute α, then λ.
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Answer: D. 0.12
α = 5; λ = (1+√26)/2 = 3.05; GE = (4/6) / (π √3.05) = 0.667 / 5.49 = 0.12.
83. The exit gradient at the downstream end of a weir floor is critical, and piping begins, when it is nearly
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At this gradient the upward force equals the submerged weight of soil.
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Answer: A. 1.0
The critical exit gradient is (G−1)/(1+e), about 1.0; a factor of safety of 5 to 7 gives safe values of 1/5 to 1/7.
84. The safe exit gradient adopted in design is highest for
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Coarser soils tolerate larger gradient.
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Answer: B. shingle (gravel)
Khosla's safe gradients: shingle 1/4–1/5, coarse sand 1/5–1/6, fine sand 1/6–1/7; coarser soils resist piping better.
85. Khosla's corrections to the uplift pressure at key points of a pile are due to
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All three relate to geometry.
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Answer: A. pile interference, floor thickness and floor slope
The standard-profile results are corrected for interference between neighbouring piles, for finite floor thickness and for floor slope.
86. The cut-off (sheet pile) at the downstream end of a weir floor is mainly intended to
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Think about seepage exit.
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Answer: B. reduce the exit gradient and prevent piping
The downstream pile confines the exit of seepage and flattens the gradient; the upstream pile mainly reduces uplift pressure under the floor.
87. The thickness of the downstream impervious floor is greatest
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Net uplift is largest at that location.
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Answer: C. in the jump zone, where uplift is highest and the cushion least
Uplift pressure is high immediately downstream of the crest while the downstream water cover is low in the jump zone, so floors are thickest there and tapered towards the tail.
88. The Lacey scour depth below the high flood level for Q = 2000 m³/s and silt factor f = 1.0 is R = 0.47 (Q/f)^(1/3) =
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Take the cube root of 2000.
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Answer: D. 5.92 m
R = 0.47 × 2000^(1/3) = 0.47 × 12.6 = 5.92 m.
89. The Lacey regime waterway (wetted perimeter) for a flood of 3600 m³/s is nearly
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Use P = 4.75√Q.
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Answer: A. 285 m
P = 4.75 √3600 = 4.75 × 60 = 285 m.
90. The discharge over a broad-crested weir of length 40 m under a head of 1.5 m, taking Q = 1.705 L H^(3/2), is nearly
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1.5^1.5 = 1.837.
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Answer: A. 125 m³/s
Q = 1.705 × 40 × 1.5^1.5 = 1.705 × 40 × 1.837 = 125 m³/s.
91. Water enters a stilling basin at depth 0.5 m and velocity 10 m/s. The sequent depth of the hydraulic jump is nearly
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Find the Froude number first.
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Answer: C. 2.95 m
Fr₁ = 10/√(9.81×0.5) = 4.52; y₂ = (y₁/2)(√(1+8Fr₁²) − 1) = 0.25 × (12.81 − 1) = 2.95 m.
92. The energy loss in the same hydraulic jump (y₁ = 0.5 m, y₂ = 2.95 m), ΔE = (y₂ − y₁)³/(4y₁y₂), is nearly
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Cube the difference of depths.
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Answer: D. 2.5 m
ΔE = (2.95 − 0.5)³ / (4 × 0.5 × 2.95) = 14.7 / 5.9 = 2.5 m.
93. A hydraulic jump with an incoming Froude number of 1.4 is classified as
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Low Froude number gives a wavy surface.
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Answer: A. an undular jump
Fr₁ of 1–1.7 gives an undular jump; 1.7–2.5 weak; 2.5–4.5 oscillating; 4.5–9 steady; above 9 strong.
94. In a stilling basin, the end sill and baffle blocks mainly serve to
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They help the jump form in a shorter distance.
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Answer: C. stabilise the jump and shorten the basin length required
Appurtenances such as chute blocks, baffle piers and end sills break up the flow and shorten the jump, so a smaller basin is needed.
7.4 River training works
29 questions · ACiE0704
95. The three stages of a river in its course from source to sea are, in order,
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Slope decreases downstream.
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Answer: B. mountain (torrent), alluvial plain and delta
A river is steep, erosive and carries boulders in the mountain stage, flows in meanders on alluvium in the plain stage and builds a delta before meeting the sea.
96. Meandering and braiding of channels are mostly observed in the
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The banks must be erodible.
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Answer: C. alluvial plain stage of a river
In the plain stage the river flows in its own deposited alluvium and can shift laterally, creating meanders and braids.
97. Which of the following is NOT an objective of river training?
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Pick the aim that would make flooding worse.
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Answer: D. to promote deposition of sediment so as to raise the river bed
Training aims to give a stable, efficient channel; promoting bed aggradation would raise flood levels and defeat that purpose.
98. River training works for the discharge of floods are known as
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It concerns the highest stage.
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Answer: C. high water training
High water training (levees, flood embankments, spurs) controls floods; low-water training deepens channels for navigation; mean-water training regulates the dominant discharge channel.
99. A river reach of channel length 18 km occurs along a valley length of 12 km. The sinuosity ratio is
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Channel length is divided by valley length.
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Answer: A. 1.5
Sinuosity = channel length / valley length = 18/12 = 1.5; values above about 1.5 indicate meandering.
100. A braided river is typically characterised by
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Think of Himalayan rivers leaving the hills.
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Answer: D. a wide shallow channel with many bar-split paths and heavy bed load
Braiding occurs when the bed-load exceeds the stream's capacity to carry it; deposited bars split the flow into many shifting channels.
101. Guide banks are constructed on the banks of a river at a diversion or bridge site mainly to
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They channel the river to the structure.
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Answer: C. guide the flow axially through the structure and prevent outflanking
Guide banks confine and direct the river flow through the waterway at right angles to the axis of the structure, with uniform distribution of discharge.
102. As a thumb rule, the length of a guide bank upstream of the structure is kept about
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Upstream is much longer than downstream.
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Answer: D. 1.0 to 1.5 times the waterway width, and the downstream length about 0.2 to 0.25 times
Upstream guide banks must be longer to align the approach flow; downstream banks only guide the outflow and are short.
103. Guide banks are given a curved head and tail (in plan) in order to
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Avoid abrupt change of flow direction.
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Answer: A. give a smooth transition of flow and reduce scour at the nose
A circular head (and tail) prevents flow separation and sharp attack at the nose, where maximum scour occurs.
104. The Lacey scour depth R for a river with Q = 3000 m³/s and silt factor f = 1.0 is R = 0.47 (Q/f)^(1/3). If the maximum scour at the nose of the guide bank is taken as 2R below HFL, it is nearly
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Find R first and double it.
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Answer: C. 13.6 m
R = 0.47 × 3000^(1/3) = 0.47 × 14.42 = 6.78 m; 2R = 13.6 m.
105. A launching apron is provided at the toe of a guide bank in order to
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It is a reserve of stone.
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Answer: A. protect the toe by launching into the scour hole
Stones laid flat on the river bed fall into the scour hole as the bed erodes, forming a protective sloping layer on the face of the bank.
106. A launching apron launches over a slope of 2H:1V to cover the scour depth D. The length of the launched slope is
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Use the Pythagoras theorem on the slope.
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Answer: D. √5 D = 2.24 D
On a 2H:1V slope the slope length for vertical depth D is D√(1+2²) = 2.24 D.
107. A launching apron is expected to launch on a 2H:1V slope to a scour depth of 3 m below the bed, with a thickness of 1.2 m in the launched position. The stone required per metre run is nearly
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Volume per metre = sloping length × thickness.
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Answer: A. 8.05 m³
Slope length = 3√5 = 6.71 m; volume per metre = 6.71 × 1.2 = 8.05 m³.
108. The thickness of stone pitching on guide bank slopes is commonly estimated as t = 0.06 Q^(1/3) (t in m, Q in m³/s). For Q = 1000 m³/s, t is
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The cube root of 1000 is 10.
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Answer: B. 0.6 m
t = 0.06 × 1000^(1/3) = 0.06 × 10 = 0.6 m.
109. Levees (flood embankments) are earthen banks built
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They are longitudinal structures.
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Answer: C. parallel to the river, to confine floods within the channel
Levees confine flood flows, protect the adjacent land and are commonly built of locally available earth with stone pitching on the river face.
110. A major disadvantage of confining a silt-laden river between levees is that
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What happens to the sediment that used to spread over the floodplain?
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Answer: C. silt deposits in the confined channel, raising the bed and flood levels
Because sediment cannot spill on to the floodplain, aggradation of the bed raises the flood level and the levees need to be raised repeatedly.
111. Compared with closely spaced levees, widely spaced levees
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More room means less depth.
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Answer: D. give room for floodplain storage and lower flood levels but enclose more land
Wider spacing gives lower flood stage and velocity (less scour at levees) but more land lies between the embankments.
112. A spur (groyne) is a structure built
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It projects into the river.
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Answer: A. projecting from the bank into the stream to deflect the current
Spurs reduce the velocity along the bank, protect it from erosion and promote deposition between the spurs.
113. A spur that is inclined upstream, making an acute angle with the bank on its upstream side, is called a
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It repels the flow away from the bank.
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Answer: B. repelling spur
Repelling spurs point upstream and push the current towards the centre; attracting spurs point downstream and draw the flow near the bank, so they are not suited to bank protection.
114. A permeable spur (for example, a row of piles or bamboo/wire-crate type) differs from an impermeable spur in that it
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Water flows through the structure.
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Answer: A. slows the flow behind it and promotes silting while letting some water through
Permeable spurs act by slowing the flow and causing sediment to deposit in their lee; impermeable spurs deflect the flow.
115. Spurs for bank protection are generally most needed at the
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Where is the current strongest?
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Answer: B. concave (outer) bank of a bend
The thalweg hugs the outer bank at a bend, where scour and bank erosion are greatest.
116. An artificial cut-off across a meander loop is made to
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It makes the path shorter.
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Answer: D. shorten the course, steepen the slope and lower flood levels
A cut-off straightens the river, increasing the gradient and flood-carrying capacity locally; it may cause upstream degradation and downstream aggradation.
117. A watershed (catchment) is best defined as
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Think of the area draining to one outlet.
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Answer: A. the area draining surface runoff to a common outlet
The watershed boundary (divide) encloses the area contributing runoff to a given point on a stream.
118. A watershed of 30 km² has stream channels of total length 45 km. Its drainage density is
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Length divided by area.
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Answer: B. 1.5 km/km²
Drainage density = total channel length / area = 45/30 = 1.5 km/km².
119. Using the Universal Soil Loss Equation A = R·K·LS·C·P with R = 400, K = 0.3, LS = 1.2, C = 0.2 and P = 0.5, the average annual soil loss is
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Multiply all six factors.
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Answer: D. 14.4 t/ha/yr
A = 400 × 0.3 × 1.2 × 0.2 × 0.5 = 14.4 t/ha/year.
120. In the Universal Soil Loss Equation, the factor C represents
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C stands for cover.
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Answer: C. the cover and management factor
R = rainfall erosivity, K = soil erodibility, LS = slope length and steepness, C = cover and management, P = conservation support practice.
121. Bench terracing is mainly adopted for soil and water conservation on
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Where is slope length the main problem?
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Answer: B. steep cultivated hill slopes
Terraces break a long steep slope into short level benches, reducing runoff velocity and erosion; they are widely used for hill agriculture.
122. The primary function of a check dam in a gully is to
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It checks the flow.
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Answer: A. reduce flow velocity, trap silt and arrest head-cutting
Check dams across gullies lower the local gradient, capture silt and stabilise the channel; vegetation then helps hold the deposited soil.
123. Afforestation and vegetative cover help watershed management mainly because they
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Think about raindrops hitting bare soil.
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Answer: B. intercept rain, raise infiltration and bind soil with roots
Vegetation dissipates raindrop energy and increases roughness and infiltration, which reduces peak runoff and sediment yield.
7.5 Regulating and cross-drainage structures
30 questions · ACiE0705
124. The head regulator of a canal is provided at the
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'Head' means the start of the canal.
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Answer: D. off-take of a canal from a river or parent canal, to control the inflow
A head regulator regulates the supply, controls silt entry and can shut off the canal for repair or during floods.
125. A cross regulator in a main canal is provided
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It ponds up the water in the parent canal.
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Answer: C. to pond up the parent canal downstream of an off-take to feed it
By closing its gates, the cross regulator ponds up the parent canal to the full supply level needed for the off-take, even when the parent discharge is low.
126. An escape in a canal system is used primarily to
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A safety valve for the canal.
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Answer: A. release surplus water from the canal into a natural drain
Escapes act as safety valves, removing excess water caused by heavy rain or by closing of the gates, and can also be used for scouring silt.
127. Escapes are normally provided
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They need somewhere to put the surplus.
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Answer: C. at drain junctions, above cross-drainage works and at the tail end
Escapes are located where a drain is available to take the surplus water, upstream of major aqueducts and at the tail end.
128. The two common types of escapes are
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One passes water over the top, the other through the bottom.
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Answer: D. weir (surface) escapes and sluice escapes
A weir escape spills surplus water over a crest when the canal level rises above FSL; a sluice escape discharges through gated openings at bed level and can scour silt.
129. The discharge through a free-flowing, broad-crested head regulator is Q = (2/3) Cd √(2g) L H^(3/2). For Cd = 0.6, L = 4.0 m and H = 1.2 m, Q is nearly
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Evaluate H^(3/2) first.
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Answer: B. 9.3 m³/s
Q = (2/3)(0.6)(4.429)(4.0)(1.2^1.5 = 1.3145) = 9.3 m³/s.
130. A cross regulator has an upstream cut-off 1.5 m deep, a downstream cut-off 2.0 m deep and a floor 20 m long. Using Bligh's theory with C = 15, the maximum safe differential head (taking creep length = 2d₁ + b + 2d₂) is
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H = L / C.
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Answer: B. 1.8 m
Creep length = 2(1.5) + 20 + 2(2.0) = 27 m; H = L/C = 27/15 = 1.8 m.
131. Flexibility of an irrigation outlet is defined as the ratio of the
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It compares two relative changes.
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Answer: D. fractional change in outlet discharge to fractional change in distributary depth
F = (dq/q) / (dy/y); it measures how the outlet discharge responds when the water level in the distributary changes.
132. A modular outlet (rigid module) has a flexibility of
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Discharge does not change with levels.
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Answer: A. zero
A modular outlet delivers a constant discharge irrespective of water level fluctuations in the distributary and the watercourse, so dq = 0 and F = 0.
133. An outlet for which the flexibility is equal to 1 is called
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Change in outlet discharge is in proportion with that of the distributary.
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Answer: C. proportional
With F = 1 the outlet discharge changes in the same ratio as the distributary discharge, giving an equitable distribution of water.
134. An open sluice outlet and a pipe outlet are classified as
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Does the discharge change with head?
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Answer: B. non-modular outlets, as their discharge varies with the water-level difference
Both depend on the head at the outlet, so they are non-modular (the pipe outlet is a submerged or free orifice-type outlet).
135. A pipe outlet of diameter 150 mm discharges freely like an orifice with Cd = 0.62 under a head of 0.4 m to its centre. The discharge Q = Cd·a·√(2gH) is nearly
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Orifice formula in SI units, then convert m³/s to L/s.
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Answer: D. 31 L/s
a = π(0.15)²/4 = 0.01767 m²; Q = 0.62 × 0.01767 × √(2×9.81×0.4) = 30.7 L/s.
136. A submerged pipe outlet 6 m long and 0.3 m in diameter operates under a difference of water levels of 0.3 m. Taking the entry loss coefficient 0.5, f = 0.02 and exit loss as one velocity head, the discharge is nearly
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Add all losses in terms of velocity head.
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Answer: C. 124 L/s
H = (0.5 + fL/D + 1)V²/2g → V = √(2×9.81×0.3/1.9) = 1.76 m/s; Q = V × π(0.3)²/4 = 124 L/s.
137. A pipe outlet is said to be 'submerged' when
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Look at the tail-water level.
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Answer: A. the tail water covers the pipe crown and the pipe flows full
In a submerged pipe outlet, the discharge depends on the difference between the upstream and downstream water levels, not the depth of water over the pipe.
138. A drop (fall) is provided in a canal when
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Think of the ground sloping more than the bed.
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Answer: A. ground slope exceeds the permissible canal bed slope
Falls lower the canal bed in steps and dissipate the excess energy so that the bed slope remains within the non-scouring limit.
139. A vertical drop fall is generally recommended for a fall height up to about
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It is for small drops.
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Answer: A. 1.5 m
Vertical drop falls are economical for small drops (up to roughly 1.5 m); larger drops need glacis or rapid-type falls.
140. Water passes over the crest of a vertical drop fall with q = 3.0 m³/s per metre width. The critical depth over the crest is nearly
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Use yc = (q²/g)^(1/3).
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Answer: A. 0.97 m
yc = (q²/g)^(1/3) = (9/9.81)^(1/3) = 0.97 m.
141. A vertical drop fall with a head loss of 1.2 m is founded on fine micaceous sand (Bligh's C = 15). The minimum total creep length of the impervious floor is
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Bligh's L = C × H.
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Answer: B. 18 m
L = C H = 15 × 1.2 = 18 m.
142. The cistern at the base of a vertical drop fall serves to
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Where does the energy get destroyed?
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Answer: D. provide a water cushion in which the jump forms
The depressed cistern keeps the tailwater above the sequent depth so that the jump forms within the floor.
143. An aqueduct is a cross-drainage structure in which
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The canal is the upper channel.
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Answer: D. the canal is carried in a trough over the drainage, whose HFL is below the trough
In an aqueduct the canal is conveyed in a trough above the HFL of the drainage, which flows freely under the arches.
144. A syphon aqueduct differs from an aqueduct in that in a syphon aqueduct
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The barrels carry drain water under pressure.
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Answer: A. the drain HFL is above the trough bottom, so barrels run full
The drainage bed is below the canal bed but its HFL exceeds the trough bottom, so the barrels run full under siphonic action.
145. A super-passage is a cross-drainage work in which
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The drain is in the trough.
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Answer: B. the drainage is carried over the canal, whose FSL is below the trough bed
A super-passage is the reverse of an aqueduct: the drain is above the canal, and the canal flows with a free surface below it.
146. In a canal siphon, the
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The canal goes under the drain.
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Answer: B. canal flows under pressure in barrels below the drainage bed
A canal siphon is the reverse of a syphon aqueduct: the drain is above the canal and the canal flows full under pressure in barrels beneath the drain.
147. A level crossing is a cross-drainage work that is provided when
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Both are at the same level.
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Answer: C. the canal and drain are at about the same level, with regulators to control flow
At a level crossing, a crest wall and regulators let the drainage flow through the canal section or take surplus water out of the canal.
148. A canal is to cross a drainage whose HFL is 1.5 m below the canal bed and whose bed is well below the canal bed. The most suitable work is an
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The drain does not wet the trough.
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Answer: B. aqueduct
When the canal bed is well above the HFL of the drainage, the canal is carried over it on an aqueduct with free flow in both.
149. The canal waterway of an aqueduct is often contracted (flumed) because
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A narrower trough costs less.
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Answer: B. a narrower trough saves cost and higher velocity is allowed
A flumed trough reduces the length and width of the aqueduct; the canal section is gradually transitioned to the narrower trough.
150. Dickens' formula Q = C A^(3/4) (A in km², Q in m³/s) is used to estimate the flood discharge of a drainage. For C = 14 and A = 16 km², Q is nearly
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16^(3/4) = 8.
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Answer: D. 112 m³/s
Q = 14 × 16^(0.75) = 14 × 8 = 112 m³/s.
151. Ryves' formula Q = C A^(2/3) with C = 6.8 and A = 27 km² gives a flood discharge of
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27^(2/3) = 9.
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Answer: A. 61.2 m³/s
Q = 6.8 × 27^(2/3) = 6.8 × 9 = 61.2 m³/s.
152. A syphon aqueduct carries a drainage flood of 60 m³/s through 3 barrels, each 4 m wide and 3 m high, running full. The mean velocity in the barrels is nearly
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Area of all barrels together.
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Answer: C. 1.67 m/s
V = Q/A = 60 / (3 × 4 × 3) = 1.67 m/s.
153. The type of cross-drainage work selected at a site depends mainly on
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It is a question of levels.
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Answer: C. the relative levels of the canal and the drainage (beds, FSL and HFL)
Whether the canal passes over, under or level with the drain is decided by comparing canal bed/FSL with drainage bed/HFL, together with the drainage discharge and the cost.
7.6 Water logging and drainage
30 questions · ACiE0706
154. Waterlogging of agricultural land occurs when
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Think of the water table and the root zone.
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Answer: D. the water table rises and saturates the root zone, displacing soil air
In waterlogged soil the pores of the root zone are filled with water, so roots cannot respire and crop growth is hampered.
155. Which of the following is a major cause of waterlogging in canal commands?
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Inputs to the groundwater.
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Answer: A. seepage from unlined canals and over-irrigation of fields
Seepage from unlined canals and distributaries, and deep percolation from excess irrigation, add to the groundwater and raise the water table.
156. Obstruction of natural drainage by canals, roads and railway embankments tends to
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Water cannot flow away.
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Answer: D. cause ponding and a rise in the water table upstream
Embankments crossing natural drains hold up runoff and seepage, so inadequate cross-drainage works lead to waterlogging.
157. A harmful effect of waterlogging on crops is that
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Roots need air.
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Answer: C. root respiration and aerobic activity fall, reducing yield
Poor aeration reduces root growth and nitrification, and waterlogged soil is also colder; yields fall.
158. In waterlogged areas of arid and semi-arid regions, salts accumulate at the surface because
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Evaporation leaves the salt.
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Answer: A. water rises by capillarity and evaporates, leaving salts behind
Capillary rise from a shallow water table brings salts to the surface where evaporation deposits them, forming saline or alkaline land.
159. Lining of canals and distributaries helps to control waterlogging because it
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Cutting the inflow to groundwater.
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Answer: D. reduces seepage losses that recharge the water table
Lining cuts seepage by as much as 80–90 per cent, reducing groundwater recharge.
160. Which of the following is NOT a preventive measure against waterlogging?
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Choose the one that adds water.
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Answer: C. increasing the intensity of irrigation without improving drainage
Extra irrigation adds water to the root zone and the aquifer; the others reduce recharge or remove surplus water.
161. Pumping groundwater by tube wells in a canal command (conjunctive use) helps in preventing waterlogging because it
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It acts on the water table directly.
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Answer: C. lowers the water table and gives an extra water supply
Vertical drainage by tube wells draws down the water table and the abstracted water can supplement canal supplies if its quality is acceptable.
162. A soil is classified as saline when its saturation extract has
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Salinity is measured by EC of saturation extract.
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Answer: C. electrical conductivity above 4 dS/m, ESP below 15 and pH below 8.5
Saline soils (white alkali) have high soluble salts but little exchangeable sodium; sodic soils have ESP above 15; saline-sodic soils have both.
163. A sodic (alkali) soil is characterised by
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Sodium is the key word.
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Answer: B. exchangeable sodium percentage above 15 and pH generally above 8.5
High exchangeable sodium disperses clay and gives poor structure and permeability.
164. The chemical commonly used to reclaim sodic (alkali) soils is
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It supplies calcium.
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Answer: C. gypsum
Gypsum (CaSO₄·2H₂O) supplies calcium to replace exchangeable sodium, which is then leached away as sodium sulphate.
165. A water has Na⁺ = 10, Ca²⁺ = 4 and Mg²⁺ = 2 meq/L. Its sodium adsorption ratio, SAR = Na/√[(Ca+Mg)/2], is nearly
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Take the average of Ca and Mg first.
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Answer: B. 5.8
SAR = 10 / √3 = 5.8.
166. The irrigation water has EC = 1.2 dS/m, and the drainage water leaving the root zone is allowed EC = 6 dS/m. The leaching requirement LR = ECiw/ECdw is
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Ratio of irrigation water to drainage water salinity.
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Answer: D. 0.20
LR = 1.2/6 = 0.20, i.e. 20% of the applied water must drain through the root zone.
167. With a leaching requirement of 0.2 and crop evapotranspiration of 600 mm, the depth of irrigation water to be applied, d = ET/(1 − LR), is
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Divide by (1 − LR).
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Answer: A. 750 mm
d = 600 / (1 − 0.2) = 750 mm.
168. Surface drainage is provided mainly to
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It works at the ground surface.
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Answer: B. remove excess rain and irrigation water from the land surface
Surface drains (field drains, collector and main drains) carry away ponded water through shallow open ditches.
169. A random surface drainage system is mostly used when
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Random means irregular.
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Answer: A. the field has scattered depressions that are joined by drains
Random drains pass through individual wet spots; parallel and cross-slope layouts suit uniform land, usually combined with land grading.
170. A surface drain serves a catchment of 100 ha with runoff coefficient 0.4 and design rainfall intensity 50 mm/h. By the rational formula Q = 0.00278 C i A (A in ha, i in mm/h), the peak discharge is nearly
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Multiply all four factors.
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Answer: A. 5.56 m³/s
Q = 0.00278 × 0.4 × 50 × 100 = 5.56 m³/s.
171. A drainage coefficient of 7 mm/day is adopted for a field of 40 ha. The discharge to be removed by the subsurface drains is nearly
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Convert mm × area to m³ per day, then to L/s.
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Answer: A. 32.4 L/s
Volume per day = 0.007 × 400 000 m² = 2800 m³; Q = 2800/86 400 = 0.0324 m³/s = 32.4 L/s.
172. Subsurface drains (tile drains) are preferred to open drains in intensively cultivated land because they
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Think of land use and machinery.
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Answer: B. do not occupy productive land and do not impede farm operations
Buried pipes leave the land fully cultivable, and without a need for bridges or crossings; but their initial cost is higher.
173. Mole drains are unlined underground channels formed by a mole plough, and are suited to
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The channel has no lining, so the soil must hold itself.
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Answer: D. stable clay soils
In stable clay the mole channel keeps its shape for several years; in sandy or unstable soils it collapses.
174. In a herringbone pattern of subsurface drainage, the lateral drains
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The shape is in the name.
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Answer: D. join the main drain obliquely from both sides, like the bones of a fish
Herringbone systems suit narrow depressions or valleys where the main drain runs along the lowest line and laterals enter from both sides.
175. A gravel or synthetic envelope around a perforated subsurface drain pipe is provided to
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It is a filter.
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Answer: B. keep soil particles out of the pipe while letting water in
The envelope acts as a filter and also improves flow convergence near the drain.
176. Hooghoudt's steady-state spacing is L² = (8Kdh + 4Kh²)/q. For K = 1 m/day, equivalent depth d = 2.0 m, h = 0.8 m and q = 0.005 m/day, the drain spacing is nearly
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Evaluate L² first and take the square root.
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Answer: A. 55 m
L² = (8×1×2×0.8 + 4×1×0.64)/0.005 = (12.8 + 2.56)/0.005 = 3072; L = 55 m.
177. Donnan's formula for drains laid on an impervious layer is L² = 4Kh²/q. With K = 1 m/day, h = 0.8 m and q = 0.005 m/day, the drain spacing is nearly
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Take the square root at the end.
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Answer: A. 22.6 m
L² = 4 × 1 × 0.64 / 0.005 = 512; L = 22.6 m.
178. If the hydraulic conductivity of the soil is increased four times, with other factors unchanged, the spacing of subsurface drains by Hooghoudt/Donnan formulae becomes
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L varies with the square root of K.
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Answer: C. two times
L ∝ √K, so a four-fold increase in K doubles the spacing.
179. The equivalent depth d in Hooghoudt's equation is introduced to account for
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Flow lines converge near the drain.
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Answer: B. extra head loss from convergent flow near the drain
Hooghoudt replaced the actual depth to the impervious layer with a smaller equivalent depth to account for the convergence of flow lines near the drain.
180. A circular subsurface drain of diameter 0.2 m with n = 0.012 is laid at a slope of 0.002. Using Q = (0.3117/n) D^(8/3) S^(1/2) for full flow, the discharge is nearly
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Express Q in m³/s first.
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Answer: D. 15.9 L/s
Q = (0.3117/0.012) × 0.2^(2.667) × 0.002^0.5 = 25.98 × 0.01367 × 0.04472 = 0.0159 m³/s.
181. Vertical drainage by tube wells is most effective for lowering the water table where
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Pumped water must flow readily to the well.
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Answer: B. the aquifer is thick and permeable and the water is usable
Tube wells can draw down large areas, and if the water is usable it supplements irrigation.
182. An interceptor drain is provided to
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It cuts off the flow coming from above.
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Answer: B. cut off seepage from higher ground before it reaches the low area
Interceptors are laid across the direction of groundwater flow at the toe of slopes or alongside canals to cut off seepage.
183. Electrical conductivity (EC) of a soil saturation extract is used to measure its
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More dissolved salts, more conduction.
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Answer: C. salinity
EC increases with the concentration of dissolved salts and is measured in dS/m (mmho/cm) as the index of salinity.