Irrigation miscellaneous


  1. A field channel has cultivable commanded area of 2000 hectares. The intensities of irrigation for gram and wheat are 30% and 50% respectively. Gram has a kor period of 18 days, kor depth of 12 cm, while wheat has a kor period of 18 days and a kor depth of 15 cm. The discharge (in m³/s) required in the field channel to supply water to the commanded area during the kor period is ____









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    1.427

    Q =
    A
    D

    Duty, D = 8.64
    B
    Δ

    B – kor period (days)
    Δ – kor depth (m)
    Gram
    QG =
    AG
    =
    2000 × 0.3
    = 0.463 m³
    DG8.64 ×
    18
    0.12

    Wheat
    QW =
    AW
    =
    2000 × 0.5
    = 0.964 m³
    DW8.64 ×
    18
    0.15

    Discharge required = Q1 + Q2 = 1.427 m³/s

    Correct Option: B

    1.427

    Q =
    A
    D

    Duty, D = 8.64
    B
    Δ

    B – kor period (days)
    Δ – kor depth (m)
    Gram
    QG =
    AG
    =
    2000 × 0.3
    = 0.463 m³
    DG8.64 ×
    18
    0.12

    Wheat
    QW =
    AW
    =
    2000 × 0.5
    = 0.964 m³
    DW8.64 ×
    18
    0.15

    Discharge required = Q1 + Q2 = 1.427 m³/s


  1. A rectangular channel of 2.5 m width is carrying a discharge of 4 m³/s. Considering that acceleration due to gravity as 9.81 m/s², the velocity of flow (in m/s) corresponding to the critical depth (at which the specific energy is minimum) is ________.









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    2.45 to 2.55

    c =
    g

    q =
    Q
    =
    4
    = 1.6m³/s/m
    B2.5

    ∴ yc =
    (1.6)²
    = 0.639 m
    9.81

    At critical depth,
    ∴ yc =
    Vc
    2

    =
    0.639
    2g2

    ∴ V = √0.639 × g = 2.5 m/s

    Correct Option: C

    2.45 to 2.55

    c =
    g

    q =
    Q
    =
    4
    = 1.6m³/s/m
    B2.5

    ∴ yc =
    (1.6)²
    = 0.639 m
    9.81

    At critical depth,
    ∴ yc =
    Vc
    2

    =
    0.639
    2g2

    ∴ V = √0.639 × g = 2.5 m/s



  1. The transplantation of rice requires 10 days and total depth of water required during transplantation is 48 cm. During transplantation, there is an effective rainfall (useful for irrigation) of 8 cm. The duty of irrigation water (in hectares/cumec) is









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    D = 864 B/Δ hectares/cumecs

    =
    864 × 10
    = 216 hectares/cumecs
    (48 – 8) × 10

    Correct Option: B

    D = 864 B/Δ hectares/cumecs

    =
    864 × 10
    = 216 hectares/cumecs
    (48 – 8) × 10


  1. Wheat crop requires 55 cm of water during 120 days of base period. The total rainfall during this period is 100 mm. Assume the irrigation efficiency to be 60%. The area (in ha) of the land which can be irrigated with a canal flow of 0.01 m³/s is









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    Actual water depth required (D1) = 55 – total rainfall
    = 55 – 10 = 45 cm

    D1 =
    864 × B
    D

    ⇒ 45 =
    864 × 120
    D

    D = 2304 ha/cumec
    1 m³/s discharge → 2304 hectares (irrigation)
    ∴ 0.01 m³/s discharge → 2304 × 0.01 = 23.04 hectares
    Actual irrigated area = 23.04 ×
    60
    = 13.824 hectares.
    100

    Correct Option: A

    Actual water depth required (D1) = 55 – total rainfall
    = 55 – 10 = 45 cm

    D1 =
    864 × B
    D

    ⇒ 45 =
    864 × 120
    D

    D = 2304 ha/cumec
    1 m³/s discharge → 2304 hectares (irrigation)
    ∴ 0.01 m³/s discharge → 2304 × 0.01 = 23.04 hectares
    Actual irrigated area = 23.04 ×
    60
    = 13.824 hectares.
    100



  1. An agricultural land of 437 ha is to be irrigated for a particular crop. The base period of the crop is 90 days and the total depth of water required by the crop is 105 cm. If a rainfall of 15 cm occurs during the base period, the duty of irrigation water is









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    Net water required = 105 – 15 = 90 cm
    D × Delta = Duty × Base period

    D =
    1 × (90 × 24 × 60 × 60)
    × 104 (m² to hectare) = 864 hectares/curve
    0.9

    Duty =
    864 × B
    =
    864 × 90
    = 864 ha/cumec
    Δ90

    Correct Option: D

    Net water required = 105 – 15 = 90 cm
    D × Delta = Duty × Base period

    D =
    1 × (90 × 24 × 60 × 60)
    × 104 (m² to hectare) = 864 hectares/curve
    0.9

    Duty =
    864 × B
    =
    864 × 90
    = 864 ha/cumec
    Δ90