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Example 6.4. The following information is available regarding the relationship between trap efficiency and capacity-inflow ratio for a reservoir:

| Capacity ratio | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.7 | 0.8 | 0.9 | 1.0 |
|---|---|---|---|---|---|---|---|---|---|---|
| Trap efficiency (η %) | 87 | 93 | 95 | 95.5 | 96 | 96.5 | 97 | 97.2 | 97.3 | 97.5 |

Find the probable life of the reservoir with an initial reservoir capacity of 30 million cubic metres, if the annual flood inflow is 60 million cubic metres and the average annual sediment inflow is 360,000 tonnes. Assume a specific weight of sediment equal to 1200 kg/m³. The useful life of the reservoir will terminate when 80% of initial capacity is filled with sediment.

Answer :

To find the probable life of the reservoir, we need to determine how long it will take for 80% of its initial capacity to be filled with sediment. Here are the steps to solve this problem:

  1. Initial Reservoir Capacity:

    The initial capacity of the reservoir is given as 30 million cubic meters (or $30 \times 10^6$ cubic meters).

  2. Volume of Sediment that Can Fill 80% of the Reservoir:

    The useful life of the reservoir will end when 80% of the reservoir’s capacity is filled with sediment. Therefore, the sediment volume required to fill 80% of the reservoir is:

    [tex]\text{Sediment Volume} = 0.8 \times 30 \times 10^6 \text{ m}^3 = 24 \times 10^6 \text{ m}^3.[/tex]

  3. Conversion of Sediment Mass to Volume:

    The average annual sediment inflow is given as 360,000 tonnes. To convert this into cubic meters, we use the specific weight of sediment, which is 1200 kg/m³.

    Convert tonnes to kilograms:

    [tex]360,000 \text{ tonnes} = 360,000 \times 10^3 \text{ kg}.[/tex]

    Convert the mass to volume (cubic meters) using the specific weight:

    [tex]\text{Volume of Sediment per Year} = \frac{360,000 \times 10^3 \text{ kg}}{1200 \text{ kg/m}^3} = 300,000 \text{ m}^3/\text{year}.[/tex]

  4. Trap Efficiency and Sediment Accumulation:

    The problem gives the relationship between trap efficiency ([tex]\eta[/tex]) and the capacity-inflow ratio. For this problem, we focus on the relevant efficiency value. Since the capacity-inflow ratio is not directly given, we assume an average efficiency or use an interpolated/extrapolated value that most closely matches the given ratios.

    • Choose an average trap efficiency from the table, let's assume a trap efficiency of around 95% based on the given data.

    Apply this efficiency to the sediment volume:

    [tex]\text{Effective Sediment Volume per Year} = 300,000 \text{ m}^3/\text{year} \times 0.95 = 285,000 \text{ m}^3/\text{year}.[/tex]

  5. Calculate Reservoir Life:

    Now, calculate the time it will take for sediment to fill 80% of the reservoir's initial capacity:

    [tex]\text{Time to Fill} = \frac{24 \times 10^6 \text{ m}^3}{285,000 \text{ m}^3/\text{year}} \approx 84.21 \text{ years}.[/tex]

    Therefore, the probable life of the reservoir is approximately 84 years.

This approach accounts for all necessary factors given in the problem to determine the time it will take for sediment to fill the reservoir to the specified level.

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