- 2074 Chaitra · 8 marks
Balancing reservoir, intermittent supply 7-10 and 17-20
Short answer
Daily demand is 20 m³ supplied intermittently 6 h/day at 3.33 m³/h. Mass-curve gives maximum surplus 2.00 m³ and maximum deficit 4.00 m³, so balancing capacity = 6.00 m³ (30% of daily demand).
A rural area has a design year demand of water 20,000 litres per day. The demand is met by an intermittent system of supply two times a day at 7-10 and 17-20 (altogether 6 hrs).
| Time (Hour) | 5-7 | 7-12 | 12-17 | 17-20 | 20-5 |
|---|---|---|---|---|---|
| Consumption (%) | 20 | 35 | 10 | 25 | 10 |
Determine the balancing reservoir capacity for that rural area.
Given data
- Daily demand = 20,000 L = 20 m³/day
- Intermittent supply in 7–10 h and 17–20 h (6 h in all): inflow rate = 20/6 = 3.33 m³/h
- Demand: 5–7 h 20%, 7–12 h 35%, 12–17 h 10%, 17–20 h 25%, 20–5 h 10%
Mass-curve table (analytical method)
Inflow: 3.33 m³/h during 7–10 h and 17–20 h only. 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 | 4.00 | 0.00 | -4.00 | -4.00 |
| 07:00–10:00 | 4.20 | 10.00 | +5.80 | +1.80 |
| 10:00–12:00 | 2.80 | 0.00 | -2.80 | -1.00 |
| 12:00–17:00 | 2.00 | 0.00 | -2.00 | -3.00 |
| 17:00–20:00 | 5.00 | 10.00 | +5.00 | +2.00 |
| 20:00–05:00 | 2.00 | 0.00 | -2.00 | +0.00 |
Total demand = 20.00 m³; total inflow = 20.00 m³.
Capacity
- Maximum cumulative surplus = 2.00 m³
- Maximum cumulative deficit = 4.00 m³
- Balancing capacity = max. surplus + max. deficit = 2.00 + 4.00 = 6.00 m³
Answer: Balancing reservoir capacity = 6 m³ (30% of the daily demand).
✓ Answer checked by Claude Opus 5.5 (high effort)
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