- 2075 Chaitra · 8 marks
Service reservoir capacity, 150,000 population
Short answer
Daily demand is 16,500 m³ and pumping rate is 2,750 m³/h for 6 h. Mass-curve gives maximum surplus 3,300 m³ and maximum deficit 3,300 m³, so service reservoir capacity = 6,600 m³ (40% of daily demand).
Water is to be supplied to a municipality in Nepal with forecasted population of 150,000 with 110 litres per capita per day. The variation in water demand is mentioned below. Calculate the capacity of service reservoir considering pumping at 6-9am and 6-9pm respectively. Neglect the fire demand and use analytical method.
| Time | 6am - 9am | 9am - 12 noon | 12 noon - 3pm | 3pm - 6pm | 6pm - 9pm | 9pm - 6am |
|---|---|---|---|---|---|---|
| % Demand | 30 | 10 | 10 | 20 | 25 | 5 |
Given data
- Population 150,000 × 110 lpcd = 16,500 m³/day
- Pumping 6–9 h and 18–21 h (6 h), at a rate of 16,500/6 = 2,750 m³/h
- Fire demand neglected
Mass-curve table (analytical method)
Inflow: 2,750 m³/h during 6–9 h and 18–21 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 |
|---|---|---|---|---|
| 06:00–09:00 | 4950.00 | 8250.00 | +3300.00 | +3300.00 |
| 09:00–12:00 | 1650.00 | 0.00 | -1650.00 | +1650.00 |
| 12:00–15:00 | 1650.00 | 0.00 | -1650.00 | 0.00 |
| 15:00–18:00 | 3300.00 | 0.00 | -3300.00 | -3300.00 |
| 18:00–21:00 | 4125.00 | 8250.00 | +4125.00 | +825.00 |
| 21:00–06:00 | 825.00 | 0.00 | -825.00 | 0.00 |
Total demand = 16500.00 m³; total inflow = 16500.00 m³.
Capacity
- Maximum cumulative surplus = 3300.00 m³
- Maximum cumulative deficit = 3300.00 m³
- Balancing capacity = max. surplus + max. deficit = 3300.00 + 3300.00 = 6600.00 m³
Answer: Service reservoir capacity = 6,600 m³ (40% of the daily demand).
✓ Answer checked by Claude Opus 5.5 (high effort)
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