--- title: "Reservoir storage for wheat, rice, maize" subject: "Irrigation and Drainage Engineering" chapter: 2 key: "semester_6/irrigation_and_drainage_engineering:reservoir-supply-40000ha-wheat-rice-maize" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/reservoir-supply-40000ha-wheat-rice-maize site: BCT Engineering --- # Reservoir storage for wheat, rice, maize Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements - Asked in: 2076 Baisakh, 8 marks - Video: [Unit 1 |Storage capacity of reservoir problem | irrigation Engineering](https://www.youtube.com/watch?v=yLYmWdBRUuE) — Dr.Brindha Sathiaseelan - 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/reservoir-supply-40000ha-wheat-rice-maize - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2 ## Question A reservoir has to supply irrigation water to 40,000 hectares land. Calculate the storage required in the reservoir to meet the irrigation demand of various crops as detailed below: | Crop | Base period (days) | Duty (ha/cumecs) | Intensity of irrigation | |---|---|---|---| | Wheat | 120 | 2400 | 20% | | Rice | 140 | 1000 | 15% | | Maize | 100 | 1800 | 20% | ## Short answer Area = intensity × 40000 ha. Q = A/D. Compute volume = Q × B × 86400 for wheat (34.56 Mm³), rice (72.58 Mm³), maize (38.40 Mm³). Sum is 145.54 Mm³. No losses given. Storage required ≈ 145.54 million m³. ## Answer **Given:** CCA = 40,000 ha. Area of each crop = intensity x CCA. No canal or reservoir losses are given, so the duties are taken as measured at the reservoir (canal head); losses are neglected. **Method:** $Q = A/D$ and $V = Q\times B\times 86400$. | Crop | Area (ha) | Duty (ha/cumec) | Q = A/D (cumec) | B (days) | Volume (Mm$^3$) | |---|---|---|---|---|---| | Wheat | 8,000 | 2,400 | 3.333 | 120 | 34.560 | | Rice | 6,000 | 1,000 | 6.000 | 140 | 72.576 | | Maize | 8,000 | 1,800 | 4.444 | 100 | 38.400 | $$ V_{total} = 145.536\times10^6\ \text{m}^3 $$ **Answer: storage required in the reservoir $\approx$ 145.54 million m$^3$.** (If canal and reservoir losses were to be included, the volume would be divided by $(1-\text{loss})$ for each.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the crop areas found for the 40,000 ha command? A. Intensity x CCA: wheat 8000, rice 6000, maize 8000 ha B. Duty x CCA for each crop C. Equal thirds of 40,000 ha each D. Base period x intensity Answer: A. Wheat 20%, rice 15%, maize 20% of 40000 ha. (Key point: Crop areas) 2. No losses are given. How does the answer treat them? A. A 25% loss is assumed by default B. Losses are neglected and duties are taken as measured at the canal head/reservoir C. Duty is converted using 8.64 D. Losses are added to the base period Answer: B. The problem gives no loss data, so volume = Q x B x 86400 directly. (Key point: Treatment of losses) 3. What is the storage needed for rice (6000 ha, duty 1000, base period 140 days)? A. 43.2 million m^3 B. 36.3 million m^3 C. 72.58 million m^3 D. 120.9 million m^3 Answer: C. Q = 6000/1000 = 6 cumec; 6 x 140 x 86400 = 72.576 Mm^3. (Key point: Rice volume) 4. How is the total storage obtained and what is it? A. Largest crop volume only, 72.58 million m^3 B. Average of the three volumes, 48.5 million m^3 C. Sum divided by 0.85, 171 million m^3 D. Sum of the three crop volumes, about 145.54 million m^3 Answer: D. 34.560 + 72.576 + 38.400 = 145.536 Mm^3. 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