--- title: "Impounded reservoir, 6202 million L per month" subject: "Water Supply Engineering" chapter: 2 key: "semester_5/water_supply_engineering:impounded-reservoir-c69" source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/impounded-reservoir-c69 site: BCT Engineering --- # Impounded reservoir, 6202 million L per month Water Supply Engineering · Chapter 2: Sources of Water - Asked in: 2069 Chaitra, 4 marks - Video: [Unit:2 | Numerical from Impounded Reserviour | Water Supply Engineering | Prashant YT | 5th Sem |](https://www.youtube.com/watch?v=OqX0FOaZxHk) — Prashant YT - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/impounded-reservoir-c69 - Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-2 ## Question The city has an average water demand of 6202 million liters per month. Calculate the capacity of the impounded reservoir. The flow in the river is shown below. | Month | Inflow (m$^3$/s) | |---|---| | January | 2.97 | | February | 1.99 | | March | 1.00 | | April | 0.00 | | May | 0.51 | | June | 1.00 | | July | 2.00 | | August | 3.00 | | September | 4.00 | | October | 5.00 | | November | 4.00 | | December | 2.80 | ## Short answer Monthly inflows are converted to volumes. The analytical mass curve method finds the largest cumulative deficit of 20.40×10^6 m³ at the end of July. Required reservoir capacity ≈ 20.40 million m³ (20,405 ML). ## Answer Convert the monthly flows to volumes ($V = Q\times86400\times\text{days}$) and apply the mass curve (analytical) method. Demand $D = 6202\ \text{ML/month} = 6.202\ \text{Mm}^3$ per month (1 ML = 1000 m$^3$). #### Monthly inflow volumes | Month | Days | Inflow (m3/s) | Volume (Mm3) | |---|---|---|---| | Jan | 31 | 2.97 | 7.955 | | Feb | 28 | 1.99 | 4.814 | | Mar | 31 | 1.00 | 2.678 | | Apr | 30 | 0.00 | 0.000 | | May | 31 | 0.51 | 1.366 | | Jun | 30 | 1.00 | 2.592 | | Jul | 31 | 2.00 | 5.357 | | Aug | 31 | 3.00 | 8.035 | | Sep | 30 | 4.00 | 10.368 | | Oct | 31 | 5.00 | 13.392 | | Nov | 30 | 4.00 | 10.368 | | Dec | 31 | 2.80 | 7.500 | Total inflow $= 74.42$ Mm$^3$; total demand $= 12\times6.202 = 74.42$ Mm$^3$. The two are equal, so the reservoir must carry the whole yearly regulation. #### Cumulative deficit (reservoir full at the start of the deficit period) | Month | Inflow (Mm3) | Demand (Mm3) | Surplus/deficit | Cumulative deficit | |---|---|---|---|---| | Feb | 4.814 | 6.202 | -1.388 | -1.388 | | Mar | 2.678 | 6.202 | -3.524 | -4.911 | | Apr | 0.000 | 6.202 | -6.202 | -11.113 | | May | 1.366 | 6.202 | -4.836 | -15.949 | | Jun | 2.592 | 6.202 | -3.610 | -19.559 | | Jul | 5.357 | 6.202 | -0.845 | -20.405 | The inflow is below the demand from Feb to Jul; the cumulative deficit is largest at the end of Jul. **Answer: Capacity of the impounded reservoir = 20.40 million m$^3$ ($\approx 20{,}405$ ML).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is a monthly inflow volume obtained from a flow in m3/s? A. V = Q x 86400 x number of days in the month B. V = Q x 24 x 30 only C. V = Q / 86400 D. V = Q x 1000 Answer: A. Seconds in a day are 86400, multiplied by days in the month. (Key point: Unit conversion) 2. What is 6202 million litres per month in Mm3? A. 62.02 Mm3 B. 6.202 Mm3 C. 0.6202 Mm3 D. 6202 Mm3 Answer: B. 1 ML = 1000 m3, so 6202 ML = 6.202 million m3. (Key point: Demand conversion) 3. Over which months is the cumulative deficit built up? A. Aug to Dec B. Jan to Apr C. Feb to Jul D. Sep to Nov Answer: C. Inflow is below 6.202 Mm3 from Feb to Jul. (Key point: Deficit period) 4. What is the reservoir capacity? A. 11.11 Mm3 B. 15.95 Mm3 C. 19.56 Mm3 D. 20.40 Mm3 Answer: D. The maximum cumulative deficit occurs at end of July and is 20.405 Mm3. 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