This set of Irrigation Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Design Capacity for an Irrigation Canal”.
1. Kor demand of the crop should be taken into account while fixing the capacity of a canal.
Explanation: Suppose take wheat crop for example, let us say it requires a 60 cm of water in a total period of about 240 days (for a given area of field) thus giving an average outlet factor of 3456 hectares/cumec. But the kor depth of wheat is 14 cm in about 4 weeks which gives an outlet of 1728 hectares/cumec. Therefore it is clear that for a given area (A) outlet factor of 1728 is very much more than 3456. Hence kor demand should be given importance while designing irrigation canal.
2. What is the most important point to be considered while fixing the canal capacity?
a) Keenest Demand
b) Average Demand
c) Water Demand
Explanation: While fixing the capacity of a canal the main important that should be kept in mind is the keenest demand, but not the average demand. For instance, let rice require 160 cm of water during 320 days which gives an outlet factor of 1728 hectares/cumec. (i.e D = 864B/∆ = 864 x 320 / 160 = 1728).
3. Let us consider in a given area the plantation of a certain crop takes 20 days, and the total water depth required by this crop is 80 cm on the field. Find the duty of irrigation water required for the crop during this period. (a) Assuming 20% losses of water in the water courses, find duty at the head of the course. (b) Find the duty of the water at the head of the distributary, assuming 10% losses from the distributary head.
a) 182.9 hectares/cumec, 194.4 hectares/cumec
b) 172.8 hectares/cumec, 185.4 hectares/cumec
c) 172.8 hectares/cumec, 194.4 hectares/cumec
d) 185.6 hectares/cumec, 184.6 hectares/cumec
Explanation: Total depth of water needed = 80 cm
Period for which water is needed = 20 days
Duty of irrigation water = (864B/d) = (864 x 20 / 80) = 216 hectares/cumec
(a) Given 20% losses in the watercourse
= 216 x 0.8 = 172.8 hectares/cumec
(b) Given 10% losses in the head of the distributary
= 216 x 0.9 = 194.4 hectares/cumec.
4. A pump was installed in a field to supply water to the crops. The duty for this crop is 432 hectares/cumec on the field and the efficiency of pump is 50%. The sown area of the field is 5 hectares. Determine the maximum output required (H.P) of the pump, if the highest water level is 4 meters below the highest portion of the field. Assume negligible field channel losses.
Explanation: Area of field to be irrigated = 5 hectares
Duty of water for crop = 432 hectares/cumec
Discharge required for the crop is = (5 / 432) = 1/86.4 cumec
Volume of water lifted per second = 1 / 86.4 cumec
Therefore, weight of water lifted per second = (1 / 86.4) x 9.81 = 0.1135 KN/sec
(unit wt. of water = 9.81 KN/m3)
Maximum static lift of pump = 4 metres
Work done by the pump in lifting water = 0.1135 x 4 = 0.454 KWatt
The input of the pump = (0.454 / 0.735) = 0.62
(1 metric H.P = 0.735 KWatt)
Output H.P of the pump = (input/η) = (0.62 / 0.5) = 0.31 H.P.
5. At a certain place, the transplantation of a crop takes 15 days, and the total depth of water required by the crop is 90 cm on the field. During the plantation rain falls and about 20 cm is utilised to fulfil the demand. Now determine the duty of irrigation water required during the plantation.
Explanation: Total depth of water required = 90 cm
Useful rainfall = 20 cm
Extra water depth needed after useful rainfall = 90 – 20 = 70 cm
Period for which water is needed = 15 days
Duty of irrigated water (∆) = (864B/d) = (864 x 15 / 70) = 185.14 hectares/cumec.
6. Let us take the gross commanded area of a watercourse is 2000 hectares, 80% of which is culturable irrigable. Intensities of sugarcane and paddy are 30% and 50%. The duties of the crops at the head of the watercourse are 850 hectares/cumec and 1900 hectares/cumec. Find the discharge required at the head of the watercourse?
a) 0.98 cumec
b) 0.97 cumec
c) 0.9 cumec
d) 0.986 cumec
Explanation: Given G.C.A = 2000 hectares
Now the C.C.A = 2000 x 80 / 100 = 1600 hectares
Intensity of irrigation for sugarcane = 30%
Area to be irrigated under sugarcane = 1600 x 30/100 = 480 hectares
Intensity of irrigation of paddy = 50%
Area to be irrigated under paddy = 1600 x 50/100 = 800 hectares
Duty of sugarcane = 850 hectares/cumec
Duty of paddy = 1900 hectares/cumec
Discharge for sugarcane = 480 / 850 = 0.565 cumec
Discharge for paddy = 800 / 1900 = 0.421 cumec
Now the total canal capacity at the head of watercourse is = 0.565 + 0.421 = 0.986 cumec.
7. We can find out monthly or fortnightly water requirements of various crops using intervals.
Explanation: The water depth will be multiplied by crop area which is required in this interval, so as to give the volume of water required in this interval. By dividing the volume by interval, we can find out the discharge needed for each interval. The summation of this will give the discharge required for all the crops in each interval. The canal can then be designed for the maximum of these values.
8. By how much percentage canal capacity is increased to meet peak demands?
a) 30 to 35%
b) 20 to 25%
c) 15 to 20%
d) 20 to 30%
Explanation: In order to be more precise, the intervals are kept as small as possible. Monthly water requirements studies are conducted to provide a sufficient quantity of water for irrigation. So, the canal capacity is increased by 20 to 25% to meet the peak demands in a month.
9. Suppose the culturable commanded for a distributary is 8000 hectares, and the intensity of irrigation for Kharif season is 60% and that of rabi season is 30%. The average duty at the head of a distributary is 4000 hectares/cumec for Kharif season and for Rabi season it is 1800 hectares/cumec, find out discharge required at the head of the distributary?
a) 1.35 cumec
b) 1.32 cumec
c) 1.33 cumec
d) 1.3 cumec
Explanation: Area to be irrigated in Kharif season = 8000 x 60/100 = 4800 hectares
Area to be irrigated in Rabi season = 8000 x 30/100 = 2400 hectares
Water required at the head of distributary to irrigate Kharif area = 4800 / 4000 = 1.2 cumec
Water required at the head of distributary to irrigate Rabi area = 2400 / 1800 = 1.33 cumec
Therefore, the required discharge is maximum of the two = 1.33 cumec.
10. Determine the discharge required at the head of the distributary in a canal where Kharif area to be irrigated is 2400 hectares and Rabi area to be irrigated is 1800 hectares, for fulfilling maximum crop requirement. Assume suitable values for kor depth and kor period.
a) 2.08 cumec
b) 2 cumec
c) 2.1 cumec
d) 2.2 cumec
Explanation: Now let us assume kor period of 4 weeks for Rabi and 3 weeks for Kharif crop. Also, assume kor depth of 12.5 cm for Rabi and 21 cm for Kharif crop.
Outlet factor for rabi = (864B/∆) = 864 x 4 x 7 / 12.5 = 1935.36 hectares/cumec
Outlet factor for kharif = (864B/∆) = 864 x 3 x 7 / 21 = 864 hectares/cumec
Area to be irrigated for Rabi season = 2400 hectares
Area to be irrigated for Kharif season = 1800 hectares
Water required for Kharif season at the distributary = 1800 / 864 = 2.08 cumec
Water required for Rabi season at the distributary = 2400 / 1935.36 = 1.24 cumec
Therefore, the canal capacity at the head of the distributary is the maximum of the two, 2.08 cumec.
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