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Balancing reservoir for 0.25 MLD town
Short answer
Using mass curve analysis, the maximum cumulative deficit is 82 m³ and surplus 96.22 m³. Required balancing reservoir capacity is 178 m³ (0.178 ML).
For the water supply of a town with a daily requirement of 0.25 MLD, it is proposed to construct a distribution reservoir. The consumption pattern is as follows:
| Time (hr.) | Consumption |
|---|---|
| 7.00-8.00 | 30% of days supply |
| 8.00-17.00 | 35% of days supply |
| 17.00-18.30 | 30% of days supply |
| 18.30-7.00 | 5% of days supply |
The pumping is to be done at a constant rate of 0.032 ML/hr for 8.00 to 16.00 hours. Determine the required capacity of balancing reservoir using analytical method.
Given data
- Daily requirement = 0.25 MLD = 250 m³/day
- Pumping at 0.032 ML/h = 32 m³/h from 8:00 to 16:00 (8 h), so pumped volume = 256 m³
- Demand: 7–8 h = 30% (75 m³), 8–17 h = 35% (87.5 m³), 17–18.5 h = 30% (75 m³), 18.5–7 h = 5% (12.5 m³)
Hourly demand rates
- 18.5–7 h (12.5 h): 12.5/12.5 = 1.0 m³/h
- 7–8 h: 75 m³/h
- 8–17 h (9 h): 87.5/9 = 9.72 m³/h
- 17–18.5 h (1.5 h): 75/1.5 = 50 m³/h
Mass-curve table (analytical method)
| Period (h) | Pumped in (m³) | Demand (m³) | Net (m³) | Cumulative (m³) |
|---|---|---|---|---|
| 0–7 | 0 | 7.00 | −7.00 | −7.00 |
| 7–8 | 0 | 75.00 | −75.00 | −82.00 |
| 8–16 | 256.00 | 77.78 | +178.22 | +96.22 |
| 16–17 | 0 | 9.72 | −9.72 | +86.50 |
| 17–18.5 | 0 | 75.00 | −75.00 | +11.50 |
| 18.5–24 | 0 | 5.50 | −5.50 | +6.00 |
The cumulative at 24 h is +6.0 m³, which is the small excess (256 − 250) of pumping over demand.
Capacity
- Maximum cumulative deficit = 82.00 m³ (at 8:00)
- Maximum cumulative surplus = 96.22 m³ (at 16:00)
- Balancing capacity = 82.00 + 96.22 = 178.22 m³
Answer: Required balancing reservoir capacity ≈ 178 m³ (0.178 ML), i.e. about 71% of the daily demand. The tank must hold 82 m³ at midnight so that the 7–8 h peak can be met before pumping starts.
✓ Answer checked by Claude Opus 5.5 (high effort)
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