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The lowest portion of the capacity-elevation curve of a proposed irrigation reservoir draining 20 km2 of catchment is represented by the following data: i. The rate of silting for the catchment = 300 m3 / km2 / year ii. Life of the reservoir = 50 years iii. Dead storage = 30 iv. The FSD of the canal at the head = 80 cm v. The crop water requirement = 250 ha.m vi. Dependable yield of the catchment = 0.29 m Calculate the gross capacity of the reservoir.

The lowest portion of the capacity-elevation curve of a proposed irrigation reservoir draining 20 km2 of catchment is represented by the following data: i. The rate of silting for the catchment = 300 m3 / km2 / year ii. Life of the reservoir = 50 years iii. Dead storage = 30 iv. The FSD of the canal at the head = 80 cm v. The crop water requirement = 250 ha.m vi. Dependable yield of the catchment = 0.29 m Calculate the gross capacity of the reservoir. Correct Answer 317.5 ha.m

Net water demand = 250 ha.m and Reservoir losses = 15% x 250 = 37.5 ha.m Live storage to meet the given demand = 250 + 37.5 = 287.5 ha.m Gross storage required to meet the demand = live storage + dead storage = 287.5 + 30 = 317.5 ha.m Dependable yield = 0.29 x 20 x 106 = 580 ha.m The gross capacity is fixed at the lesser value of the gross storage and the dependable yield. Hence, the reservoir capacity = 317.5 ha.m.

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