Irrigation miscellaneous


  1. 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
    Δ0.2592

    Discharge, Q =
    A
    =
    100,00
    = 20 cumec
    D5000

    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
    Δ0.2592

    Discharge, Q =
    A
    =
    100,00
    = 20 cumec
    D5000

    cumec Canal is designed for more than 20 cumec to account for capanty factor and time factor.


  1. 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, ηa

    ηa =
    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, ηa

    ηa =
    water stored in root zone
    water supplied to field

    =
    2080
    × 100 = 72.22%
    2800



  1. 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 = Qwheat
    ARΔR
    =
    AWΔW
    8.64BR8.64BW

    1200 × 1.34
    =
    AW × 0.52
    140120

    ∴ AW = 2650 hectares

    Correct 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 = Qwheat
    ARΔR
    =
    AWΔW
    8.64BR8.64BW

    1200 × 1.34
    =
    AW × 0.52
    140120

    ∴ AW = 2650 hectares


  1. A tube well having a capacity of 4 m³ /hour operates for 20 hours each day during the irrigation season. How much area can be commanded if the irrigation interval is 20 days and depth of irrigation is 7 cm?









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    Total volume of water = Discharge of pump × pumping time per day × irrigation interval.
    = 40 × 20 × 20 = 1600 m³

    Area that can be commanded =
    Volume of water
    =
    1600
    Depth of water0.07

    = 22857 m² = 2.29 × 104

    Correct Option: D

    Total volume of water = Discharge of pump × pumping time per day × irrigation interval.
    = 40 × 20 × 20 = 1600 m³

    Area that can be commanded =
    Volume of water
    =
    1600
    Depth of water0.07

    = 22857 m² = 2.29 × 104



  1. 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