- 2073 Shrawan · 8 marks
Service reservoir capacity, 10,000 m3/day city
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.
The water demand of a city is 10,000 m/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.
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 of the reservoir is not given, so the level is storage ÷ . For example, if the tank has 4 m water depth for 3,167 m³, 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).
✓ Answer checked by Claude Opus 5.5 (high effort)
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