Question 1 |

Identify the cross-drainage work in the figure.

Super passage | |

Aqueduct | |

Siphon aqueduct | |

Level crossing |

Question 1 Explanation:

In given CDW, river is above the canal with underside of trough sufficiently above the full supply level of canal. Hence it is super passage.

Question 2 |

A canal supplies water to an area growing wheat over 100 hectares. The duration between the first and last watering is 120 days, and the total depth of water required by the crop is 35 \mathrm{~cm}. The most intense watering is required over a period of 30 days and requires a total depth of water equal to 12 \mathrm{~cm}. Assuming precipitation to be negligible and neglecting all losses, the minimum discharge (in \mathrm{m}^{3} / \mathrm{s}, rounded off to three decimal places) in the canal to satisfy the crop requirement is _____

0.012 | |

0.025 | |

0.046 | |

0.082 |

Question 2 Explanation:

{Area}(A)=100 ha

Total duration of irrigation =120 days

Total depth of water required =35 \mathrm{~cm}

Intense irrigation (A)

Time \left(B_{A}\right)=30 days

Depth of water \left(\Delta_{A}\right)=12 \mathrm{~cm}

{Duty}\left(D_{A}\right)=\frac{8.64 B_{A}}{\Delta_{A}}

\mathrm{D}_{\mathrm{A}}=\frac{8.64 \times 30}{0.12}

D_{A}=2160 \mathrm{ha} / \text { cumec }

Discharge required,

\begin{aligned} & Q_{A}=\frac{A}{D_{A}} \\ & Q_{A}=\frac{100}{2160}=0.0463 \mathrm{~m}^{3} / \mathrm{s} \end{aligned}

Remaining irrigation (B)

Duration \left(B_{B}\right)=90 days

Depth of water \left(\Delta_{\mathrm{B}}\right)=23 \mathrm{~cm}

\begin{aligned} \text { Duty } \left(D_{B}\right) & =\frac{8.64 \times 90}{0.23} \\ D_{B} & =3380.87 \mathrm{ha} / \text { cumec } \end{aligned}

Discharge required \left(Q_{B}\right)=\frac{A}{D_{B}}

\Rightarrow \quad \mathrm{Q}_{\mathrm{B}}=0.0296 \mathrm{~m}^{3} / \mathrm{s}

Hence, minimum capacity required =0.0463 \mathrm{~m}^{3} / \mathrm{s}.

Total duration of irrigation =120 days

Total depth of water required =35 \mathrm{~cm}

Intense irrigation (A)

Time \left(B_{A}\right)=30 days

Depth of water \left(\Delta_{A}\right)=12 \mathrm{~cm}

{Duty}\left(D_{A}\right)=\frac{8.64 B_{A}}{\Delta_{A}}

\mathrm{D}_{\mathrm{A}}=\frac{8.64 \times 30}{0.12}

D_{A}=2160 \mathrm{ha} / \text { cumec }

Discharge required,

\begin{aligned} & Q_{A}=\frac{A}{D_{A}} \\ & Q_{A}=\frac{100}{2160}=0.0463 \mathrm{~m}^{3} / \mathrm{s} \end{aligned}

Remaining irrigation (B)

Duration \left(B_{B}\right)=90 days

Depth of water \left(\Delta_{\mathrm{B}}\right)=23 \mathrm{~cm}

\begin{aligned} \text { Duty } \left(D_{B}\right) & =\frac{8.64 \times 90}{0.23} \\ D_{B} & =3380.87 \mathrm{ha} / \text { cumec } \end{aligned}

Discharge required \left(Q_{B}\right)=\frac{A}{D_{B}}

\Rightarrow \quad \mathrm{Q}_{\mathrm{B}}=0.0296 \mathrm{~m}^{3} / \mathrm{s}

Hence, minimum capacity required =0.0463 \mathrm{~m}^{3} / \mathrm{s}.

Question 3 |

In the context of cross-drainage structures, the correct statement(s) regarding
the relative positions of a natural drain (stream/river) and an irrigation canal,
is/are

In an aqueduct, natural drain water goes under the irrigation canal, whereas in a super-passage, natural drain water goes over the irrigation canal. | |

In a level crossing, natural drain water goes through the irrigation canal. | |

In an aqueduct, natural drain water goes over the irrigation canal, whereas in a super-passage, natural drain water goes under the irrigation canal. | |

In a canal syphon, natural drain water goes through the irrigation canal. |

Question 3 Explanation:

In Aqueduct and Syphon aqueduct under irrigation
canal or canal passes over natural drain.

In Super passage and Canal syphon or simply syphon, natural drain passes over irrigation canal or canal goes under natural drain.

In level crossing, natural drain water goes through the irrigation canal.

In Super passage and Canal syphon or simply syphon, natural drain passes over irrigation canal or canal goes under natural drain.

In level crossing, natural drain water goes through the irrigation canal.

Question 4 |

During a particular stage of the growth of a crop, the consumptive use of water is
2.8 mm/day. The amount of water available in the soil is 50 % of the maximum
depth of available water in the root zone. Consider the maximum root zone depth
of the crop as 80 mm and the irrigation efficiency as 70%.

The interval between irrigation (in days) will be _________. (round off to the nearest integer)

The interval between irrigation (in days) will be _________. (round off to the nearest integer)

14 | |

45 | |

58 | |

78 |

Question 4 Explanation:

MTA-Marks to ALL

Frequency = \frac{\text{Depth of water needed}}{\text{Consumptive use}}=\frac{40}{2.8}=14.28 days

Frequency = \frac{\text{Depth of water needed}}{\text{Consumptive use}}=\frac{40}{2.8}=14.28 days

Question 5 |

A reservoir with a live storage of 300 million cubic metre irrigates 40000 hectares ( 1 hectare= 10^{4} \mathrm{~m}^{2}) of a crop with two fillings of the reservoir. If the base period of the crop is 120 days, the duty for this crop (in hectares per cumec, round off to integer) will then be ________

691 | |

127 | |

985 | |

456 |

Question 5 Explanation:

\begin{aligned} \text { Live storage } &=300 \mathrm{Mm}^{3} \\ \text { Area } &=40000 \text { hectare }\\ \text {Since 2 filling,} & \text{ so volume of water needed,} \\ &=600 \mathrm{Mm}^{3}\\ B&=120 days \\\text { Duty } &=\frac{8.64 B}{\Delta} \\ \Delta &=\frac{600 \times 10^{6}}{40000 \times 10^{4}}=1.5 \mathrm{~m} \\ \text { Duty } &=\frac{8.64 \times 120}{1.5}=691.2 \text { ha/cumec } \end{aligned}

There are 5 questions to complete.