--- title: "Reservoir capacity with intensity table" subject: "Irrigation and Drainage Engineering" chapter: 2 key: "semester_6/irrigation_and_drainage_engineering:reservoir-capacity-40000ha-table-78p" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/reservoir-capacity-40000ha-table-78p site: BCT Engineering --- # Reservoir capacity with intensity table Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements - Asked in: 2078 Poush, 6 marks - Video: [Calculation of Reservoir Capacity : Numerical problem : Duty and Delta](https://www.youtube.com/watch?v=D6v4aQ60sZY) — The Classroom - 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-capacity-40000ha-table-78p - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2 ## Question The base period, the intensity of irrigation and duty of water for various crops under the canal system are given. Determine the reservoir capacity if the cultivable command area is 40,000 ha, canal losses are 25% and reservoir losses are 15%. | Crop | Base Period (Days) | Duty of water at the field (ha/m³/sec) | Irrigation Intensity (%) | |---|---|---|---| | Wheat | 120 | 1,800 | 20 | | Sugarcane | 360 | 1,700 | 20 | | Cotton | 180 | 1,400 | 10 | | Rice | 120 | 800 | 15 | | Vegetable | 120 | 700 | 15 | ## Short answer Total water volume required at field is about 403.5 million m³. After adding 25% canal losses and 15% reservoir losses, the required reservoir capacity is about 633 million m³. ## Answer **Given:** CCA = 40,000 ha, canal loss = 25%, reservoir loss = 15%. Area of each crop = intensity x CCA (wheat 8000 ha, sugarcane 8000 ha, cotton 4000 ha, rice 6000 ha, vegetable 6000 ha). **Method.** For each crop, discharge needed at the field $Q = A/D$; volume for the base period $V = Q\times B\times 86400$ m$^3$. The losses are taken as a fraction of the water entering that part of the system, so field volume is divided by $(1-\text{loss})$. | Crop | Area (ha) | Duty (ha/cumec) | Q = A/D (cumec) | B (days) | Volume = Q x B x 86400 (Mm$^3$) | |---|---|---|---|---|---| | Wheat | 8,000 | 1,800 | 4.444 | 120 | 46.080 | | Sugarcane | 8,000 | 1,700 | 4.706 | 360 | 146.372 | | Cotton | 4,000 | 1,400 | 2.857 | 180 | 44.434 | | Rice | 6,000 | 800 | 7.500 | 120 | 77.760 | | Vegetable | 6,000 | 700 | 8.571 | 120 | 88.869 | **Total volume at the field:** $V_f = 403.515\times10^6$ m$^3$ **Allow canal loss (25%):** $$ V_{canal\ head} = \frac{403.515}{1-0.25} = 538.019\times10^6\ \text{m}^3 $$ **Allow reservoir loss (15%):** $$ V_{reservoir} = \frac{538.019}{1-0.15} = 632.964\times10^6\ \text{m}^3 $$ **Answer: reservoir capacity $\approx$ 632.96 million m$^3$ (0.6330 Gm$^3$).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the area of each crop found? A. Irrigation intensity x CCA B. Base period x CCA C. Duty x CCA D. CCA / intensity Answer: A. E.g. wheat 20% x 40000 = 8000 ha. (Key point: Crop area) 2. How is the volume for each crop calculated? A. V = A x D x B B. Q = A/D, then V = Q x B x 86400 C. V = Q / (B x 86400) D. V = A/D x B only in cumec-days without converting Answer: B. Discharge at field times base period in seconds gives volume in m^3. (Key point: Volume per crop) 3. How is the 25% canal loss allowed for? A. Multiply the field volume by 0.75 B. Add 25 million m^3 C. Divide the field volume by (1 - 0.25) D. Divide by 1.25 Answer: C. Losses are a fraction of water entering the canal, so field volume/0.75 = 538.0 Mm^3. (Key point: Canal loss) 4. What is the final step and result? A. Multiply 538.0 by 0.85, giving 457 Mm^3 B. Add 15% to the field volume 403.5, giving 464 Mm^3 C. Take 538.0 Mm^3 as the capacity D. Divide canal-head volume by (1 - 0.15), giving about 632.96 Mm^3 Answer: D. The reservoir must release the canal-head volume plus its own losses: 538.019/0.85 = 632.96 Mm^3. 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