Q21GATE 2004MCQ

A sprinkler irrigation system is suitable when

Q22GATE 2004MCQ

A canal irrigates a portion of a culturable command area of grow sugarcane and wheat. The average discharges required to grow sugarcane and wheat are, respectively, 0.36 and 0.27 cumecs. The time factor is 0.9. The required design capacity of the canal is

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📖 Explanation

Design discharge $=\frac{0.36+0.27}{0.9}=0.70$

Q23GATE 2003MCQ

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 1.1 m. The consumptive use of crop at this stage is 2.5 mm/day and there is no precipitation during this period. Irrigation efficiency is 65%. It is intended to raise the moisture content to the field capacity in 8 days through irrigation. The necessary depth of irrigation is

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📖 Explanation

Depth of water required to raise the moisture content to the field capacity (without considering loss due to evaporation).

\begin{array}{l} =1.5 \times 1.1 \times (0.15-0.05) \\ =0.165 \mathrm{m} \end{array}

Adding, consumptive use @ 2.5mm / day for 8 days. $\text { Water required }=0.165+\frac{8 \times 2.5}{1000}=0.185 \mathrm{m}$ Irrigation efficiency =65 % $\therefore$ The necessary depth of irrigation $=\frac{185}{0.65} \mathrm{mm}=284.62 \mathrm{mm}$

Q24GATE 2003MCQ

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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📖 Explanation

Given Delta, $\Delta=25.92 \mathrm{cm}=0.2592 \mathrm{m}$ Base period, B=150 days Then Duty, $D=\frac{8.64 B}{\Delta}=\frac{8.64 \times 150}{0.2592}$ $\Rightarrow \quad D=5000 \text{ha/cumec}$ $\therefore$ Minimum discharge $=\frac{100,000}{5000}$ $=20 \text{cumecs}$ But the distributary channel should be designed for a discharge more than 20 cumecs.

Q25GATE 2002MCQ

A canal was designed to supply the irrigation needs of 1200 hectares to land growing rice of 140 days base period having a Delta of 134 cms. If this canal water is used to irriagate wheat of base period 120 days having a Delta of 52 cm, the area (in Hectares) that can be irrigtated is

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📖 Explanation

\begin{aligned} Q&=\frac{1200 \times 10^{4} \times 134}{100 \times 140}\\ &=\frac{\text { Area }}{120} \times \frac{52}{100} \\ \text { Area } &=2650 \mathrm{ha} \end{aligned}

Q26GATE 2001MCQ

A field was supplied water from an irrigation tank at a rate of 120 lit/sec 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 lit/sec. The runoff loss in the field was estimated as 800 $m^3$ . The application efficiency in this situation is

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📖 Explanation

Water application efficiency,

\eta_{\mathrm{a}}=\frac{\begin{array}{c}\text{Water stored \in the}\\ \text { root zone during irrigation }\end{array}}{\text { Water delivered to the field }} \times 100

Water supplied to field during 8 hour @ 100 lit/s

\begin{array}{l} =100 \times 8 \times 60 \times 60 \\ =2880 \mathrm{m}^{3} \end{array}

Runoff loss in the field $=800 \mathrm{m}^{3}$ Water stored in the root zone

\begin{aligned} &=2880-800=2080 \mathrm{m}^{3} \\ \therefore \quad \eta_{\mathrm{a}}&=\frac{2080}{2880} \times 100=72.22 \% \end{aligned}