- 2072 Chaitra · 4 marks
Impounded reservoir capacity, 4x10^6 m3 demand
Short answer
Using the mass curve method, the maximum cumulative deficit of inflow below the constant monthly demand of 4×10^6 m³ is 9.5×10^6 m³ at the end of March. The required storage capacity is 9.5 million m³.
Determine the storage capacity of the impounded reservoir for a city with a water demand of per month. The run-off discharge in the river is given in the table:
| Month | April | May | June | July | Aug. | Sep. | Oct. | Nov. | Dec. | Jan. | Feb. | March |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Inflow () | 5.1 | 5.9 | 8.3 | 8.9 | 7.5 | 5.0 | 3.5 | 3.0 | 2.0 | 1.6 | 2 | 2.4 |
The storage of an impounded reservoir must supply the demand during the period when inflow is less than the demand. It is found by the analytical (mass curve) method: take demand /month and find the largest cumulative deficit. All volumes are in million m.
Check: total inflow and total demand , so the river can meet the demand over the year.
The reservoir is assumed full at the end of September, when the deficit period starts.
| Month | Inflow (Mm3) | Demand (Mm3) | Surplus/deficit | Cumulative deficit |
|---|---|---|---|---|
| Oct | 3.50 | 4.00 | -0.50 | -0.50 |
| Nov | 3.00 | 4.00 | -1.00 | -1.50 |
| Dec | 2.00 | 4.00 | -2.00 | -3.50 |
| Jan | 1.60 | 4.00 | -2.40 | -5.90 |
| Feb | 2.00 | 4.00 | -2.00 | -7.90 |
| Mar | 2.40 | 4.00 | -1.60 | -9.50 |
The inflow is below the demand from October to March, and the cumulative deficit reaches its maximum at the end of March.
Answer: Storage capacity of the impounded reservoir = 9.5 million m ().
✓ Answer checked by Claude Opus 5.5 (high effort)
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