Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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}
Question 4 |
Crops are grown in a field having soil, which has field capacity of 30% and permanent
wilting point of 13%. The effective depth of root zone is 80 cm. Irrigation water is supplied
when the average soil moisture drops to 20%. Consider density of the soil as 1500 kg/m^3
and density of water as 1000 kg/m^3. If the daily consumptive use of water for the crops
is 2 mm, the frequency of irrigating the crops (in days), is
7 | |
10 | |
11 | |
13 |
Question 4 Explanation:
As per GATE Official Answer Key MTA (Marks to All)
FC=30%
PWP=13%
\begin{aligned} d_w&=\frac{\gamma _d}{\gamma _w}d \times (FC-OMC) \\ &= \frac{1500}{1000}\times 80 (0.3-0.2)\\ &= 12cm=120mm \end{aligned}
Consumptive use = 2 mm/day
So, frequency of irrigation =120/2 = 60day
FC=30%
PWP=13%
\begin{aligned} d_w&=\frac{\gamma _d}{\gamma _w}d \times (FC-OMC) \\ &= \frac{1500}{1000}\times 80 (0.3-0.2)\\ &= 12cm=120mm \end{aligned}
Consumptive use = 2 mm/day
So, frequency of irrigation =120/2 = 60day
Question 5 |
The data for an agricultural field for a specific month are given below:
Pan Evaporation = 100 mm
Effective Rainfall = 20 mm (after deducting losses due to runoff and deep percolation)
Crop Coefficient = 0.4
Irrigation Efficiency = 0.5
The amount of irrigation water (in mm) to be applied to the field in that month, is
Pan Evaporation = 100 mm
Effective Rainfall = 20 mm (after deducting losses due to runoff and deep percolation)
Crop Coefficient = 0.4
Irrigation Efficiency = 0.5
The amount of irrigation water (in mm) to be applied to the field in that month, is
0 | |
20 | |
40 | |
80 |
Question 5 Explanation:
Water required by the crop = 100 x 0.4 = 40mm
Effective rainfall =20mm
Additional water required =20mm
Amount of water required after accounting irrigation efficiency =\frac{20}{0.5}=40mm
Effective rainfall =20mm
Additional water required =20mm
Amount of water required after accounting irrigation efficiency =\frac{20}{0.5}=40mm
There are 5 questions to complete.