# Irrigation Engineering

 Question 1
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
 A In an aqueduct, natural drain water goes under the irrigation canal, whereas in a super-passage, natural drain water goes over the irrigation canal. B In a level crossing, natural drain water goes through the irrigation canal. C In an aqueduct, natural drain water goes over the irrigation canal, whereas in a super-passage, natural drain water goes under the irrigation canal. D In a canal syphon, natural drain water goes through the irrigation canal.
GATE CE 2022 SET-1      Design of Stable Channels
Question 1 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. 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)
 A 14 B 45 C 58 D 78
GATE CE 2022 SET-1      Crop Water Requirements
Question 2 Explanation:
MTA-Marks to ALL 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 ________
 A 691 B 127 C 985 D 456
GATE CE 2021 SET-2      Crop Water Requirements
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
An unlined canal under regime conditions along with a silt factor of 1 has a width of flow 71.25m. Assuming the unlined canal as a wide channel, the corresponding average depth of flow (in m, round off to two decimal places) in the canal will be ______________
 A 1.25 B 4.58 C 3.82 D 2.94
GATE CE 2021 SET-1      Design of Stable Channels
Question 4 Explanation: R = D (for wide rectangular channel)
\begin{aligned} &\begin{aligned} A f^{2} &=140\left(\frac{2}{5} f R\right)^{5 / 2} \\ (B D) f^{2} &=140\left(\frac{2}{5} f \times D\right)^{5 / 2} \\ (71.25 \times D) \times 1 &=140\left(\frac{2}{5} \times 1 \times D\right)^{5 / 2} \\ D \times 0.5089 &=\left(\frac{2}{5}\right)^{5 / 2} \times(D)^{5 / 2} \\ D^{3 / 2} &=5.029 \\ \text { or } \qquad \qquad \qquad \qquad D &=2.94 \mathrm{~m} \end{aligned} \end{aligned}
 Question 5
A concrete dam holds 10 m of static water as shown in the figure (not drawn to the scale). The uplift assumed to vary linearly from full hydrostatic head at the heel, to zero at the toe of dam. The coefficient of friction between the dam and foundation soil is 0.45. Specific weights of concrete and water are 24 $kN/m^3$ and 9.81 $kN/m^3$, respectively. For NO sliding condition, the required minimum base width B (in m, round off to two decimal places) is __________.
 A 12.45 B 18.26 C 14.12 D 15.87
GATE CE 2020 SET-2      Design and Construction of Gravity Dams
Question 5 Explanation: \begin{aligned} \mu &=0.45\\ \gamma_{conc} &=24 kN/m^3\\ B_{min. sliding}&=\frac{10}{0.45(2.4-1)}\\ &=15.873m \end{aligned}
 Question 6
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
 A 7 B 10 C 11 D 13
GATE CE 2020 SET-2      Crop Water Requirements
Question 6 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
 Question 7
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
 A 0 B 20 C 40 D 80
GATE CE 2020 SET-1      Crop Water Requirements
Question 7 Explanation:
Water required by the crop = 100 x 0.4 = 40mm
Effective rainfall =20mm
Amount of water required after accounting irrigation efficiency $=\frac{20}{0.5}=40mm$
 Question 8
At the foot of a spillway, water flows at a depth of 23 cm with a velocity of 8.1 m/s, as shown in the figure. The flow enters as an M-3 profile in the long wide rectangular channel with bed slope = $\frac{1}{1800}$ and Manning's n= 0.015. A hydraulic jump is formed at a certain distance from the foot of the spillway. Assume the acceleration due to gravity, g = 9.81 $m/s^2$. Just before the hydraulic jump, the depth of flow $y_1$ (in m, round off to 2 decimal places) is _______
 A 0.24 B 0.42 C 0.82 D 0.96
GATE CE 2019 SET-2      Theories of Seepage, Spillways and Miscellaneous
Question 8 Explanation: \begin{aligned} y &=0.23m \\ V&=8.1m/s \\ q&=Vy=0.23 \times 8.1 \\ &= 1.863 m^3/s-m\\ S_0&=\frac{1}{800} \\ D&=0.015 \end{aligned}
$y_n=$ Normal depth of flow
R = y for wide rectangular channel
By Manning's equation
\begin{aligned} q&=\frac{y_n}{n}R^{2/3}S_0^{1/2} \\ 1.863&=\frac{y_n^{5/3}}{0.015} \times \left ( \frac{1}{1800} \right )^{1/2} \\ y_n&= 1.108m \end{aligned}
$y_1$ is conjugate depth of $y_0$,
\begin{aligned} \frac{y_1}{y_n}&=\frac{-1+\sqrt{1+8Fr_n^2}}{2} \\ \because &\;\;\left ( Fr_n^2=\frac{q^2}{gy_n^3} \right ) \\ y_1&=\frac{-1+\sqrt{1+8 \times \frac{1.863^2}{9.81 \times 1.108^3}}}{2} \times 1.108\\ &=0.418m\simeq 0.42m \end{aligned}
 Question 9
A confined aquifer of 15 m constant thickness is sandwiched between two aquicludes as shown in the figure. The heads indicated by two piezometers P and Q are 55.2 m and 34.1 m, respectively. The aquifer has a hydraulic conductivity of 80 m/day and its effective porosity is 0.25. If the distance between the piezometers is 2500 m, the time taken by the water to travel through the aquifer from piezometer location P to Q (in days,round off to 1 decimal place) is______
 A 925.6 B 654.4 C 369.6 D 254.2
GATE CE 2019 SET-2      Theories of Seepage, Spillways and Miscellaneous
Question 9 Explanation:
Discharge velocity,
\begin{aligned} V&=ki\\ V&=k\frac{h_1-h_2}{L}\\ &=80 \left ( \frac{55.2-34.1}{2500} \right )\\ &=0.6752 m/day\\ &\text{Porosity}, n =0.25\\ &\therefore \;\;\text{Seepage velocity,}\\ V_s&=\frac{V}{n}=\frac{0.6752}{0.25}\\ &=2.7008 m/day\\ \text{Time taken}&=\frac{L}{V_s}\\ &=\frac{2500}{2.7008}=925.65 days \end{aligned}
 Question 10
The command area of a canal grows only one crop, i.e., wheat. The base period of wheat is 120 days and its total water requirement, $\Delta$, is 40 cm. If the canal discharge is 2 $m^3/s$, the area, in hectares, rounded off to the nearest integer, which could be irrigated (neglectingall losses) is _______
 A 5184 B 4282 C 2364 D 9852
GATE CE 2019 SET-2      Crop Water Requirements
Question 10 Explanation:
Given data:
Base period, B = 120 days
Delta of crop, $\Delta = 40 cm$
Discharge, $Q = 2 m^3/s$
Area to be irrigated, A = ?
Duty of water, $\Delta =\frac{864 \times B}{\Delta }$ ha/cumec
and Area to be irrigated; A = Q x D
$\Rightarrow A = 2 m^3/s \times \frac{864 \times 120}{40}\; ha/cumec$
$A = 5184\; ha$
There are 10 questions to complete.