--- title: "Service reservoir capacity, 10,000 m3/day city" subject: "Water Supply Engineering" chapter: 7 key: "semester_5/water_supply_engineering:res-s73" source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/res-s73 site: BCT Engineering --- # Service reservoir capacity, 10,000 m3/day city Water Supply Engineering · Chapter 7: Reservoirs and Distribution System - Asked in: 2073 Shrawan, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/res-s73 - Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7 ## Question The water demand of a city is 10,000 m$^3$/day. The water demand is to meet from the river flowing under gravity to the reservoir. The water is supplied to the consumers from the reservoir by continuous system. Calculate the capacity of service reservoir for the consumption pattern as shown in figure below. | Time | 05-07 | 07-12 | 12-17 | 17-19 | 19-05 | |---|---|---|---|---|---| | Water consumption (%) | 25 | 30 | 15 | 20 | 10 | Find the water level in the reservoir at 6, 12, 18 and 24 hours. ## Short answer With uniform gravity inflow of 416.67 m³/h and daily demand 10,000 m³, mass-curve analysis gives maximum deficit 3,166.67 m³ and no surplus. Capacity ≈ 3,167 m³; storage at 6, 12, 18, 24 h is 2,333, 583, 583 and 1,583 m³, with levels obtained by dividing storage by tank area. ## Answer #### Given data - Daily demand = 10,000 m³/day - River water enters the reservoir by gravity at a uniform rate: 10,000/24 = 416.67 m³/h - Demand: 5–7 h 25%, 7–12 h 30%, 12–17 h 15%, 17–19 h 20%, 19–5 h 10% #### Mass-curve table (analytical method) Inflow: 416.67 m³/h for 24 h. Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank. | Time | Demand (m³) | Inflow (m³) | Inflow − demand | Cumulative | |---|---|---|---|---| | 05:00–07:00 | 2500.00 | 833.33 | -1666.67 | -1666.67 | | 07:00–12:00 | 3000.00 | 2083.33 | -916.67 | -2583.33 | | 12:00–17:00 | 1500.00 | 2083.33 | +583.33 | -2000.00 | | 17:00–19:00 | 2000.00 | 833.33 | -1166.67 | -3166.67 | | 19:00–05:00 | 1000.00 | 4166.67 | +3166.67 | +0.00 | Total demand = 10000.00 m³; total inflow = 10000.00 m³. #### Capacity - Maximum cumulative surplus = 0.00 m³ - Maximum cumulative deficit = 3166.67 m³ - Balancing capacity = max. surplus + max. deficit = 0.00 + 3166.67 = **3166.67 m³** #### Water level in the reservoir The reservoir is empty at 19:00 (largest deficit −3,166.67 m³) and full (3,166.67 m³) at 05:00. Storage at any time = cumulative value + 3,166.67 m³. | Time | Storage in tank (m³) | Level (m) | |---|---|---| | 6:00 | 2,333.3 | 2,333.3/A | | 12:00 | 583.3 | 583.3/A | | 18:00 | 583.3 | 583.3/A | | 24:00 | 1,583.3 | 1,583.3/A | The plan area $A$ of the reservoir is not given, so the level is storage ÷ $A$. For example, if the tank has 4 m water depth for 3,167 m³, $A = 3167/4 = 792$ m², giving levels of 2.95 m, 0.74 m, 0.74 m and 2.00 m. **Answer: Service reservoir capacity ≈ 3,167 m³ (31.7% of the daily demand).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the inflow taken as 416.67 m3/h in every hour? A. The river supplies the daily demand at a uniform rate over 24 h (10,000/24) B. Inflow equals the peak hourly demand C. Inflow is only during high-demand hours D. Inflow is 10,000/12 because the day is split into two halves Answer: A. Continuous gravity supply from the river is uniform, so the daily volume is divided by 24. (Key point: Gravity inflow) 2. Here the maximum surplus is 0 and the largest deficit is 3,166.67 m3. What is the capacity? A. 0 m3 B. 1,583.3 m3 C. 3,166.67 m3 D. 6,333.3 m3 Answer: C. Capacity = 0 + 3,166.67 = 3,166.67 m3. (Key point: Reading the cumulative column) 3. The tank is empty at 19:00. How is storage at 12:00 found? A. Cumulative value at 12:00 + 3,166.67 m3 (capacity) B. Cumulative value at 12:00 only C. Demand in 12-17 h D. Capacity minus daily demand Answer: A. Storage = cumulative value + the deficit that sets the empty point; -2,583.33 + 3,166.67 = 583.3 m3. 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