--- title: "Duty and minor discharge, wheat 400 ha" subject: "Irrigation and Drainage Engineering" chapter: 2 key: "semester_6/irrigation_and_drainage_engineering:minor-wheat-400ha-duty-discharge" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/minor-wheat-400ha-duty-discharge site: BCT Engineering --- # Duty and minor discharge, wheat 400 ha Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements - Asked in: 2070 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/minor-wheat-400ha-duty-discharge - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2 ## Question A minor commands 400 ha of irrigable area. It is proposed to consider wheat crop in the whole command area. The kor period for the wheat is considered 3 weeks. The kor depth has been assessed to be 10 cm. In this period 2.75 cm of rainfall is normally expected with such an intensity that 50% of this could be taken as superfluous (surface runoff). Considering 10% conveyance loss find out (a) duty of the canal water at the field head and (b) discharge of the minor at upstream head. ## Short answer Effective rainfall = 0.5 × 2.75 = 1.375 cm, net kor depth = 10 – 1.375 = 8.625 cm. Duty at field head = 8.64×21/0.08625 ≈ 2104 ha/cumec. Q_field = 400/2104 = 0.1901 cumec. Adding 10% loss gives Q_head ≈ 0.211 cumec. ## Answer **Given:** area = 400 ha (all wheat), kor period $B = 3$ weeks = 21 days, kor depth = 10 cm, rainfall = 2.75 cm of which 50% is surface runoff (not useful), conveyance loss = 10%. **Effective rainfall** $$ P_e = 0.5\times 2.75 = 1.375\ \text{cm} $$ **Net kor depth to be supplied by irrigation** $$ \Delta = 10 - 1.375 = 8.625\ \text{cm} = 0.08625\ \text{m} $$ **(a) Duty at the field head** $$ D_{field} = \frac{8.64\,B}{\Delta} = \frac{8.64\times 21}{0.08625} = 2103.7\ \text{ha/cumec} $$ **(b) Discharge of the minor at the upstream head** $$ Q_{field} = \frac{400}{2103.7} = 0.1901\ \text{cumec} $$ Allowing 10% conveyance loss: $$ Q_{head} = \frac{0.1901}{1-0.10} = 0.2113\ \text{cumec}\ (211\ \text{l/s}) $$ (The duty at the head of the minor is $D_{field}\times0.9 = 1893.3$ ha/cumec.) **Answer: (a) duty at field head $\approx$ 2104 ha/cumec; (b) discharge at the head of the minor $\approx$ 0.211 cumec.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Rainfall is 2.75 cm and 50% is surface runoff. What effective rainfall is used? A. 1.375 cm B. 2.75 cm C. 0 cm D. 5.5 cm Answer: A. Only the non-runoff half is useful: 0.5 x 2.75 = 1.375 cm. (Key point: Effective rainfall) 2. What depth must irrigation supply? A. 10 + 1.375 = 11.375 cm B. 10 - 1.375 = 8.625 cm C. 10 - 2.75 = 7.25 cm D. 10 cm Answer: B. Effective rainfall reduces the kor depth to be supplied. (Key point: Net kor depth) 3. Which formula and result give the duty at the field head? A. D = 8.64 delta / B = 0.035 ha/cumec B. D = B / delta = 243 ha/cumec C. D = 8.64 B / delta = 2104 ha/cumec D. D = 8.64 B / 10 = 18.1 ha/cumec Answer: C. B = 21 days, delta = 0.08625 m, so D = 8.64 x 21/0.08625 = 2103.7. (Key point: Duty at field head) 4. How is the minor head discharge found from the field discharge of 0.1901 cumec? A. Multiply by 0.9 giving 0.171 cumec B. Divide by 1.10 giving 0.173 cumec C. Use 0.1901 unchanged D. Divide by (1 - 0.10) giving 0.211 cumec Answer: D. Conveyance loss means more must enter the minor head: 0.1901/0.9 = 0.2113 cumec. 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