Irrigation miscellaneous
- Uplift pressures at points E and D (Figure A) of a straight horizontal floor of negligible thickness with a sheet pile at downstream end are 28% and 20%, respectively. If the sheet pile is at upstream end of the floor (Figure B), the uplift pressures at points D1 and C1 are
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A + D1
Uplift pressure = 100 – (head loss along DC)
= 100 – 20 = 80%
A + C1 Uplift pressure = 100 – (head com from E to C)
= 100 – 28 = 72%Correct Option: B
A + D1
Uplift pressure = 100 – (head loss along DC)
= 100 – 20 = 80%
A + C1 Uplift pressure = 100 – (head com from E to C)
= 100 – 28 = 72%
- A launching apron is to be designed at downstream of a weir for discharge intensity of 6.5 m³/s/m. For the design of i launching aprons the scour depth is taken two times of lacey scour depth. The silt factor of the bed material is unity. If the tailwater depth is 4.4 m, the length of launching apron in the launched position is
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Laccy apron depth,
RL = 1.35 
q² 
1/2 ƒ
q = 6.3 m³/s/m.; ƒ = 1∴ RL = 1.35 6.5 1/2 = 4.7 m 1
∴ Scour depth = 2 × 4.7 m = 9.4 m
Tar/water depth = 4.4 m
∴ Length of launching apron in the launched position = 9.4 – 4.4 = 5 mCorrect Option: C
Laccy apron depth,
RL = 1.35 q² 1/2 ƒ
q = 6.3 m³/s/m.; ƒ = 1∴ RL = 1.35 6.5 1/2 = 4.7 m 1
∴ Scour depth = 2 × 4.7 m = 9.4 m
Tar/water depth = 4.4 m
∴ Length of launching apron in the launched position = 9.4 – 4.4 = 5 m
- The culturable commanded area for a distributary is 2 × 108m². The intensity of irrigation for a crop is 40%. If kor water depth and kor period for the crop are 14 cm and 4 weeks, respectively, the peak demand discharge is
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Culturable command area (CCA) = 2 × 108 m2
Area to be irrigated,A = 40 × 2 × 108 = 2 × 107 m2 100
Period, B = 4 weeks = 7 × 4 = 28 days
Depth, Δ = 14 m = 0.14 mD = 8.64 B D = 8.64 × 28 = 1728 m²/cumec 0.14 Discharge, Q = A = 8 × 107 D 1728
(Convert to m³/s)
Q = 4.63 m³/sCorrect Option: B
Culturable command area (CCA) = 2 × 108 m2
Area to be irrigated,A = 40 × 2 × 108 = 2 × 107 m2 100
Period, B = 4 weeks = 7 × 4 = 28 days
Depth, Δ = 14 m = 0.14 mD = 8.64 B D = 8.64 × 28 = 1728 m²/cumec 0.14 Discharge, Q = A = 8 × 107 D 1728
(Convert to m³/s)
Q = 4.63 m³/s
- In a cultivated area, the soil has porosity of 45% and field capacity of 38%. For a particular crop, the root zone depth is 1.0 m, the permanent wilting point is 10% and the consumptive use is 15 mm/d. If the irrigation efficiency is 60%, what should be the frequency of irrigation such that the moisture content does not fall below 50% of the maximum available moisture?
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Field capacity = Weight of water in soil (W = r × v) Weight of soil = rw × Vv (Vv is used since water is in voids) rsoil × V 0.38 = rw × η 
Porosity, η = Vv 
rsoil V ∴ rsoil = η = 0.45 rw 0.38 0.38 Available moisture = r × (field capacity – wilting point) × depth rw = 0.45 (0.38 – 0.1) × 1000 (d = 1 m = 1000 mm) = 331 mm 0.38 Readily available moisture = 331 = 165 mm 2 Number of days irrigated = 165 × 0.6 = ≈ 6 days 15 Correct Option: B
Field capacity = Weight of water in soil (W = r × v) Weight of soil = rw × Vv (Vv is used since water is in voids) rsoil × V 0.38 = rw × η Porosity, η = Vv rsoil V ∴ rsoil = η = 0.45 rw 0.38 0.38 Available moisture = r × (field capacity – wilting point) × depth rw = 0.45 (0.38 – 0.1) × 1000 (d = 1 m = 1000 mm) = 331 mm 0.38 Readily available moisture = 331 = 165 mm 2 Number of days irrigated = 165 × 0.6 = ≈ 6 days 15
- The consumptive use of water for a crop during a particular stage of growth is 2.0 mm/day. The maximum depth of available water in the root zone is 60 mm. Irrigation is required when the amount of available water in the root zone is low. Frequency of irrigation should be
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Net available moisture = 1 × 60 = 30 mm 2 Frequency of irrigation = 30 = 15 days. 2 Correct Option: B
Net available moisture = 1 × 60 = 30 mm 2 Frequency of irrigation = 30 = 15 days. 2
