This set of Irrigation Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Canal Irrigation System – Curves in Channels”.
1. A curve in channel results in _____________
a) silting on inside (convex side) and scouring on outside (concave side)
b) scouring on inside (convex side) and scouring on outside (concave side)
c) both silting and scouring on inside
d) both silting and scouring on outside
Explanation: When a channel is aligned the proposed curve should be as gentle as possible. It leads to silting on the convex side and scouring on the concave side due to the disturbance of flow caused by curves. Stone pitching is sometimes proposed to avoid erosion and scouring.
2. If the discharging capacity of the channel is less than 0.5 cumecs then, the minimum recommended value for curve radius is _________
a) 150 m
b) 100 m
c) 300 m
d) 600 m
Explanation: If the discharge is more, the curve should be gentler and should have more radius. When the capacity is less than 0.5 cumecs, the minimum recommended curve radius is 100m. If the discharge lies in the range 0.5-3 cumecs, the value is 150m.
3. The Gross Command Area (G.C.A) represents the geographical area of the Doab.
Explanation: The G.C.A includes the cultivated as well as an uncultivated area like ponds, residential areas, roads, etc. An irrigation canal lies in the area between the two drainages and can economically irrigate it. Thus, the drainages fix the boundary of the gross command of a canal system and it becomes uneconomical to use the system across the two drainages i.e boundaries.
4. What is the Annual Intensity of irrigation (AII)?
a) Gross Irrigated Area/Cultivable Command Area
b) Cultivable Command Area/Gross Irrigated Area
c) Net Irrigated Area/Cultivable Command Area
d) Cultivable Command Area/Net Irrigated Area
Explanation: The AII is the sum total of intensities of irrigation of all the seasons of the year. IT can also be defined as the percentage of CCA which may be irrigated annually. It is thus obtained by dividing the gross irrigated area by the CCA.
5. What is Gross Cropped Area?
a) Net irrigated area + Area irrigated more than once during the same year
b) Net cropped area + Area sown more than once during the same year
c) Net irrigated area + Net cropped area
d) Total area irrigated once a year + area irrigated more than once in that year
Explanation: Sometimes two crops in two seasons are grown during a particular year on the same area. Hence this area will be sown more than once during that year. When this area is added to the area which is sown only once then the value obtained is Gross Cropped Area.
6. The ratio of the actual operating period of a distributary to the crop period is called as ____________
a) capacity factor
b) time factor
c) full supply coefficient
d) nominal duty
Explanation: No distributary is allowed to operate on all the days during any crop season to check for the dangers of over-irrigation leading to waterlogging and salinity. This ratio of actual operating period of a distributary to the crop period is called time factor. It is useful in computing the design capacity of a distributary.
7. The number of hectares irrigable per cumec of the canal capacity at its head is known as ____________
a) nominal duty
b) duty on capacity
c) design full supply discharge
d) capacity factor
Explanation: Full supply coefficient is also called duty on capacity. It is the design duty at the head of the canal. It is obtained by dividing the area estimated to be irrigated during the base period with the design full supply discharge at the head of the canal.
8. The cropped area in Rabi season is usually increased.
Explanation: The water requirement during Rabi season reduces to about 2/3rd times the full supply, the capacity factor usually varies from 0.60 to 0.70 for rabi season. The cropped area is increased in order to improve this factor.
9. The Gross Command Area for a distributary is 8000 hectares 80% of which is culturable irrigable. The intensity of irrigation for Rabi season is 50% and the average duty at the head of the distributary is 2000 hectares/cumec. Determine the discharge required at the head of the distributary.
a) 1.33 cumec
b) 1.60 cumec
c) 1.40 cumec
d) 1.44 cumec
Explanation: Cultivable Command Area CCA = 8000 x 80/100 = 6400 hectares
The area to be irrigated = CCA x Intensity of irrigation = 6400 x 50/100 = 3200 hectares
The water required at the head of the distributary = 3200/2000 = 1.6 cumec.
10. The CCA of a water course is 1500 hectares. The intensity of sugarcane and wheat crops is 20% and 40% respectively. The duties for the crops at the head of the watercourse are 730 hectares/cumec and 1600 hectares/cumec respectively. Calculate the design discharge at the outlet assuming a time factor equal to 0.8.
a) 0.55 cumec
b) 1.0 cumec
c) 0.72 cumec
d) 0.92 cumec
Explanation: Area to be irrigated under sugarcane = 1500 x 20/100 = 300 hectares
Area to be irrigated under wheat = 1500 x 40/100 = 600 hectares
Discharge required for sugarcane = 300/730 = 0.410 cumec
Discharge required for wheat = 600/1800 = 0.333 cumec
The discharge required at the head of the water course = 0.410 + 0.333 = 0.743 cumec
The actual design discharge at the outlet = 0.743/0.8 = 0.92 cumec.
11. The transplantation of rice takes 18 days and the total depth of water required by the crop is 60 cm on the field. During this transplantation period, rain starts falling and about 10 cm of rain is being utilized to fulfill the rice demand. Find the duty of irrigation water required for rice assuming 25% losses of water in watercourses.
a) 233.28 hectares/cumec
b) 240 hectares/cumec
c) 230 hectares/cumec
d) 244.44 hectares/cumec
Explanation: Extra water depth required = 60 – 10 = 50 cm
Duty of irrigation water = 864 B/D = 864 x 18/50 = 311.04 hectares/cumec
Duty at the head of the watercourse = 311.04 x (1-0.25) = 233.28 hectares/cumec.
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.