This set of Irrigation Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Lining of Canal”.
1. Irrigation water is a costly commodity.
Explanation: For good quality of irrigation products and to avoid crop failures due to natural famines and scarcity of water a huge amount of capital should be invested in the irrigation projects. For example in our country for the five year plan (1997 – 2002) the capital invested in the irrigation projects is 115000 lakh crores. This says that irrigation water is a costly commodity.
2. What type of losses can be mainly avoided by lining the canals?
a) Evaporation Losses
b) Seepage Losses
c) Erosion of Canal Bed
d) Discharge Losses at Branch Canals
Explanation: If lining is not provided in the canal then irrigation water is lost in the form of percolation and absorption as seepage losses. This loss is very significant as it reduces the potential of the irrigation water, which is a costly commodity. So, in order to reduce seepage losses the lining of the canal is needed.
3. What is the name given to the land unfit for agriculture due to waterlogging?
c) Waterlogged Area
Explanation: Due to uncontrolled seepage losses the water in the canal gets percolated and the water table level in underground rises. This level rises up to or near the ground level. This renders the land unfit for agriculture as the crop roots bring up the alkali salts to the ground level. This phenomenon is known as waterlogging and the land is called thur.
4. Lining of canals can bring high areas under command.
Explanation: With the help of lining canals can be designed for both in smaller cross section and shorter in length. Steep gradients can be provided as higher velocities are allowed, and flat slopes can also be provided without silting action on lined canals. Therefore, due to these reasons the command area can be increased.
5. Design a lined canal carrying a discharge of 20 cumecs. The slope is 1 in 7000. Assume suitable values for side slopes and lining.
a) Side Slope = 1.2: 1, Depth = 4.4 m
b) Side Slope = 1.4: 1, Depth = 4.1 m
c) Side Slope = 1.5: 1, Depth = 3.77 m
d) Side Slope = 1.3: 1, Depth = 4.5 m
Explanation: Let us assume side slope as 1.5 : 1 (1.5H : 1V), the rugosity constant as 0.015.
Now let us design a triangular section.
A = y2 (θ + cotθ), P = 2y(θ + cotθ), and R = A/P = y/2
Tanθ = (1/1.5) and cotθ = 1.5 and θ = 0.59
A = y2 (0.59 + 1.5) = y2 (2.09)
P = 2 y (0.59 + 1.5) = y (4.18)
R = 0.5y
Now Manning’s equation gives
Q = 1/n x A x R2/3 x S1/2
20 = (1/0.015) x y2 (2.09) x (0.5y) 2/3 x (1/7000)1/2
y= 3.02 m
Freeboard = 0.75 m
Total depth = 3.02 + 0.75 = 3.77 m and side slope = 1.5 : 1
6. Design a lined canal to carry a discharge of 400 cumecs and slope is 1 in 6000. Take side slopes as 1 : 1 and rugosity constant as 0.012. Assume limiting velocity as 3 m/sec.
a) Total Depth = 6.3 m, Bed Width = 15 m
b) Total Depth = 6.28 m, Bed Width = 14.2 m
c) Total Depth = 6.5 m, Bed Width = 14 m
d) Total Depth = 6.25 m, Bed Width = 14.4 m
Explanation: V = 1/n x R2/3 x S1/2
3 = (1/0.012) x R2/3 x (1/6000)1/2
R = 4.7 m
Design a trapezoidal section canal.
A = y (B + yθ + y cotθ), P = B + 2yθ + 2y cotθ
Now Tanθ = 1/1, θ = 0.785 radians, and cotθ = 1
Now A = y (B + y x 0.785 + y x 1) = y(B + 1.785y), P = B + 2y(0.785) + 2y(1) = B + 2.57y
A = Q/V = 400/3 = 133.33 sq.m
100 = y (B +1.785y), B = 100/y – 1.785y
R = A/P = 133.33/P
4.7 = 133.33/B + 2.57Y
y= 5.5 m
B = 14.4 m
Total Depth = 5.5 + 0.75 = 6.25 m
Bed Width = 14.4 m.
7. Depending on what factor does the lining of a canal can increase the capacity of the canal?
a) Width of the Canal
b) Type of Flow
d) Side Slope
Explanation: The reason is very simple. The lined surface provides a smooth surface for the flow of water with minimum resistance, whereas the unlined canal provides more resistance due to vegetation, undulations, rocks etc. Moreover the velocity of flow in lined canal is high when compared with the velocity in unlined canal, as the capacity of canal is dependent on the velocity directly if velocity increases so does the capacity of the canal.
8. Mathematically, depending on what factor we can say velocity increases the capacity of the canal?
a) Coefficient of Viscosity
b) Coefficient of Capillarity
c) Coefficient of Roughness
d) Coefficient of Resistance
Explanation: Coefficient of roughness is indirectly proportional to velocity in any mathematical formula, or condition, or equation. So, therefore the value of this coefficient is more for unlined canals than the lined canals. Hence the velocity is more for lined canals than the unlined canals and therefore capacity for lined canals is more.
9. What type of major dangers can the lining of canals extinguish?
d) Seepage Losses
Explanation: Floods are catastrophic disasters in nature. If a canal is not lined or it constructed on weak foundations then there is chance of regular floods, sometimes even flash floods. In unlined canals there is constant eroding of embankments which ultimately leads to floods. So, therefore lining of canal helps to keep a check or stop the floods, which is a major danger.
10. What costs are reduced with the help of lining a canal?
a) A.R and M.O Costs
b) Irrigation Water Costs
c) Construction Costs
d) Transportation Costs
Explanation: The expenditure on Annual repair (A.R) and Maintenance (M.O) costs are reduced with the lining of a canal. The expenditure maybe due to removal of weeds and plants, minor repairs like covering cracks, leakages, and undulations in the bed, removal silt deposited on the bed. So, therefore if lining is done the expenditure (costs) can be reduced.
Sanfoundry Global Education & Learning Series – Irrigation Engineering.
To practice all areas of Irrigation Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers.