--- title: "Rural area 22000 L/day river supply (question text cut)" subject: "Water Supply Engineering" chapter: 7 key: "semester_5/water_supply_engineering:res-a76" source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/res-a76 site: BCT Engineering --- # Rural area 22000 L/day river supply (question text cut) Water Supply Engineering · Chapter 7: Reservoirs and Distribution System - Asked in: 2076 Asoj, 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-a76 - Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7 ## Question A rural area has designed year demand of water 22000 liters per day. The demand is met by a continuous system of supply from a river source with measured safe yield of 0.25 lps. The consumption pattern is as follows: | Time (Hour) | 5-7 | 7-12 | 12-17 | 17-19 | 19-5 | |---|---|---|---|---|---| | Consumption (%) | 25 | 35 | 20 | 20 | 0 | [The printed paper states no explicit question after the data; the capacity of the balancing tank is to be determined.] ## Short answer Daily demand 22 m3, river yield 21.6 m3/day (slight deficit). Assuming inflow = yield, peak rates exceed inflow. Balancing capacity by mass curve = 9.4 m3 (adopt 10 m3). ## Answer #### Given data - Daily demand = 22,000 L = 22 m³/day - River safe yield = 0.25 lps = 21.6 m³/day = 0.9 m³/h - Demand: 5–7 h 25%, 7–12 h 35%, 12–17 h 20%, 17–19 h 20%, 19–5 h 0% #### Check of source and need of storage The printed paper gives no explicit question, so the capacity of the balancing tank is found. The safe yield (21.6 m³/day) is slightly less than the demand (22 m³/day), a deficit of 0.4 m³/day (1.8%), which a tank cannot remove (it needs a small extra supply or demand control). The demand rates in 5–7 h and 7–12 h (2.75 and 1.54 m³/h) exceed the inflow of 0.9 m³/h, so the tank is necessary. #### Mass-curve table (analytical method) Inflow: 0.9 m³/h (safe yield). 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 | 5.50 | 1.80 | -3.70 | -3.70 | | 07:00–12:00 | 7.70 | 4.50 | -3.20 | -6.90 | | 12:00–17:00 | 4.40 | 4.50 | +0.10 | -6.80 | | 17:00–19:00 | 4.40 | 1.80 | -2.60 | -9.40 | | 19:00–05:00 | 0.00 | 9.00 | +9.00 | -0.40 | Total demand = 22.00 m³; total inflow = 21.60 m³. #### Capacity - Maximum cumulative surplus = 0.00 m³ - Maximum cumulative deficit = 9.40 m³ - Balancing capacity = max. surplus + max. deficit = 0.00 + 9.40 = **9.40 m³** - Here the daily inflow (21.60 m³) is 0.40 m³ **less** than the demand, so nothing overflows; the cumulative ends at −0.40 m³, and without the extra 0.40 m³/day the tank would slowly run down day after day. **Answer: Balancing tank capacity ≈ 9.4 m³ (adopt 10 m³).** If the inflow were 22/24 = 0.917 m³/h (shortage made up), the capacity would be 9.17 m³, nearly the same. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does the safe yield (21.6 m3/day) compare with demand 22 m3/day? A. It is slightly less, a shortage of 0.4 m3/day B. It is greater C. It is equal D. It is half Answer: A. A tank cannot remove this small shortage; extra supply or demand control is needed. (Key point: Source shortage) 2. What is the demand volume in 7-12 h (35% of 22 m3)? A. 7.70 m3 B. 5.50 m3 C. 4.40 m3 D. 3.20 m3 Answer: A. 0.35 x 22 = 7.70 m3. (Key point: Demand 7-12 h) 3. What is the maximum cumulative deficit? A. 6.90 m3 B. 9.40 m3 C. 3.70 m3 D. 0.40 m3 Answer: B. After 17-19 h the cumulative is -9.40 m3. (Key point: Max deficit) 4. What is the balancing capacity? A. 9.4 m3 (adopt 10 m3) B. 22 m3 C. 0.4 m3 D. 18.8 m3 Answer: A. Max surplus is 0, so capacity = 9.40 m3. 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