Irrigation miscellaneous
- A field was supplied water from an irrigation tank at a rate of 120 L/s to irrigate an area of 2.5 hectares. The duration of irrigation is 8 hours. It was found that the actual delivery at the field, which is about 4 km from the tank, was 100 L/s. The runoff loss in the field was estimated as 800 m³. The application efficiency in this situation is
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Water supplied to field in 8 hrs (at 100 L/s)
= 100 × 8 × 60 × 60 = 2880 × 10³ L = 2880 m³
Runoff loss = 800m³
∴ Water stored is root zone = 2880 – 800 = 2080 m³
Water application efficiency, ηaηa = water stored in root zone water supplied to field = 2080 × 100 = 72.22% 2800 Correct Option: B
Water supplied to field in 8 hrs (at 100 L/s)
= 100 × 8 × 60 × 60 = 2880 × 10³ L = 2880 m³
Runoff loss = 800m³
∴ Water stored is root zone = 2880 – 800 = 2080 m³
Water application efficiency, ηaηa = water stored in root zone water supplied to field = 2080 × 100 = 72.22% 2800
- A canal was designed to supply the irrigaiton needs of 1200 hectares to land growing rice of 140 days base period having a Delta of 134 cms. If this canala water is used to irrigate wheat of base period 120 days having a Delta of 52 cm, the area (in Hectares) that can be irrigated is
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Discharge, Q = A D D = 8.64 × B Δ ∴ Q = A × Δ 8.64 × B
For rice and wheat discharge is same.
Qrice = QwheatARΔR = AWΔW 8.64BR 8.64BW 1200 × 1.34 = AW × 0.52 140 120
∴ AW = 2650 hectaresCorrect Option: A
Discharge, Q = A D D = 8.64 × B Δ ∴ Q = A × Δ 8.64 × B
For rice and wheat discharge is same.
Qrice = QwheatARΔR = AWΔW 8.64BR 8.64BW 1200 × 1.34 = AW × 0.52 140 120
∴ AW = 2650 hectares
- The total irrigation depth of water, required by a certain crop in its entire growing period (150 days), is 25.92 cm. The culturable command area for a distributary channel is 100,000 hectares. The distributary channel shall be designed for a discharge.
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Duty, D = 8.64 × B = 8.64 × 150 = 5000 ha/cumec Δ 0.2592 Discharge, Q = A = 100,00 = 20 cumec D 5000
cumec Canal is designed for more than 20 cumec to account for capanty factor and time factor.Correct Option: D
Duty, D = 8.64 × B = 8.64 × 150 = 5000 ha/cumec Δ 0.2592 Discharge, Q = A = 100,00 = 20 cumec D 5000
cumec Canal is designed for more than 20 cumec to account for capanty factor and time factor.
- The moisture content of soil in the root zone of an agricultural crop at certain stage is found to be 0.05. The field capacity of the soil is 0.15. The root zone depth is 1.1m. The consumptive use of crop at this stage is 2.5 mm/day and there is no precipitation during this peirod. Irrigation efficiency is 65%. It is intended to raise the moisture contant to the field capacity in 8 days through irrigation. The necessary depth of irrigation is
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Depth of water required for root zone
dω = S.d [Fc – ω]
Assume mass specific gravity, S = 1.40
dω = 1.4 × 1.1 [0.15 – 0.05]
= 0.154 m = 154 mm
Total consumptive use of water for 8 days
= 8 × Cu = 8 × 2.5 = 20 mm
Total depth of water to be stored in root zone = 154 + 20 = 174 mm
η = 65%Total depth of water to be supplied = 174 η = 174 = 267.7 mm ≈ 285 mm 0.6 Correct Option: D
Depth of water required for root zone
dω = S.d [Fc – ω]
Assume mass specific gravity, S = 1.40
dω = 1.4 × 1.1 [0.15 – 0.05]
= 0.154 m = 154 mm
Total consumptive use of water for 8 days
= 8 × Cu = 8 × 2.5 = 20 mm
Total depth of water to be stored in root zone = 154 + 20 = 174 mm
η = 65%Total depth of water to be supplied = 174 η = 174 = 267.7 mm ≈ 285 mm 0.6
Direction: The moisture holding capacity of the soil in a 100 hectare farm is 18 cm/m, the field is to be irrigated when 50 per cent of the available moisture in the root zone is depleted. The irrigation water is to be supplied by a pump working for 10 hours a day, and water application efficiency is 75%. Details of crops planned for cultivation area are as follows:
[2010, 2 marks each] 
- The area of crop Y that can be irrigated when the available capacity of irrigation system is 40 litres/sec is
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Moisture in 0.8 m = 0.8 × 18 = 14.4 cm
Readily available moisture = 14.4 × 50 = 7.2 cm 100 Frequency of irrigation = 7.2 = 18 days. 0.4
Quantity of water pumped is 10 h @ 40 L/s for 18 days= 40 × 10-3 × 18 × 10 × 60 × 60 m³ = 34560 mm³ 0.75 = A × 104 × 7.2 = 34560 (104: hectares to m²) 100
∴ A = 27 hectares.Correct Option: D
Moisture in 0.8 m = 0.8 × 18 = 14.4 cm
Readily available moisture = 14.4 × 50 = 7.2 cm 100 Frequency of irrigation = 7.2 = 18 days. 0.4
Quantity of water pumped is 10 h @ 40 L/s for 18 days= 40 × 10-3 × 18 × 10 × 60 × 60 m³ = 34560 mm³ 0.75 = A × 104 × 7.2 = 34560 (104: hectares to m²) 100
∴ A = 27 hectares.
