A new reservoir has a capacity of 12 3 Mm And its catchment area is 400 km². The annual sediment yield from this catchment is 0.1 ha m/km² and the trap efficiency can be assumed constant at 90%. The number of years it takes for the reservoir to lose 50% of its initial capacity is, nearly A 177 years B C 77 years 17 years D 7 years
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- Q1/A reservoir has a capacity of 50 Mm³ and average annual inflow of 75 Mm³. Area of the catchment is 1000 km². Sediment yield and the specific weight of the sediments are estimated at 550 tons per km² and 1375 kg/m³ respectively. The trap efficiency may be approximated by Y=100 [1- 1 where Y is the trap efficiency in percent and X is /the capacity inflow ratio, using three increments, calculate how long will it take for the reservoir to become 60% filled with sediment? 3 65X+1A reservoir had an average surface area of 20 km² during September 2008. In that month, the mean rate of inflow is 10 m³/s, outflow is 15 m³/s, monthly rainfall is 10 cm and change in storage is 16 Mm³. Assuming the seepage losses to be 1.8 cm, estimate the evaporation in that month.Choose the typical range of values for the storage coefficient of confined aquifers a) 1-100 b) 0.1-0.3 c) 10 ^ - 12 - 10 ^ - 8 d) 10 ^ - 6 - 10 ^ - 3
- With a catchment area of 50 hectares, a small water impounding reservoir with a capacity of 800 m3 was constructed. If the annual sediment prediction is 4.26 m3/ha, what is the probable life of the reservoir before its capacity is reduced to 400 m3? Assume a sediment trap efficiency of the dam of 25%.A reservoir had an average surface area of 20 km during June 1982. In that month the mean rate of inflow=10m/s, outflow = 15 m/s, monthly rainfall = 10 cm and change in storage = 16 million m³. Assuming the seepage losses to be 1.8 cm, estimate the evaporation in that month.C- A reservoir had an average surface area of 20 km2 during June 1982. In that month the mean rate of inflow = 10 m3 /s, outflow = 15 m3 /s, monthly rainfall = 10 cm and change in storage = 16 million m3 . Assuming the seepage losses to be 1.8 cm, estimate the evaporation in that month.
- Groundwater is pumped for domestic use from an unconfi ned aquifer (water-bearing sand layer). The thickness of the clay layer above the sand layer is 20 m and its initial porosity is 40%. After 10 years of pumping, the porosity is reduced to 30%. Determine the subsidence of the clay surface.At a reservoir, the following climatic data were observed. Estimate the mean monthly evaporation depth and the total monthly volume of evaporation from the reservoir using Rohwer’s formula if the surface area of the reservoir is 3000 ha. can help me and just i know the sourse this question pleaseEstimate the constant rate of withdrawal from a 1375 ha reservoir in a month of 30 days during which the reservoir level dropped by 0.75 m in spite of an average inflow into the reservoir of 500,000 m 3 /day. During the month the average seepage loss from the reservoir was 2.5 cm, total precipitation on the reservoir was 18.5 cm and the total evaporation was 9.5 cm.
- A reservoir surface covers an area of 850 km2 and has an average depth of 18.7 m. The inflow of the reservoir is from a river with an average flow rate of 2500m3/s and an average suspended-sediment concentration of 250 mg/l. Estimate the rate at which the depth of the reservoir is decreasing due to sediment accumulation and the time it will take for the reservoir storage to decrease by 10%. Assume that the accumulated sediment has a bulk density of 1600 kg/m3.7.75. A reservoir covers an area of 850 km² and has an aver- age depth of 18.7 m. The inflow to the reservoir is from a river with an average flow rate of 2500 m³/s and an average suspended-sediment concentration of 250 mg/L. Estimate the rate at which the depth of the reservoir is decreasing due to sediment accumulation, and the time it will take for the reservoir storage to decrease by 10%. Assume that the accumulated sediment has a bulk den- sity of 1600 kg/m³.1) For hydration to occur, what is the approximate minimum w/c ratio required? A) 0.5 B) 0.15 C) 0.35 D) 0.25 2) 1 MPa (Mega Pascal) is equivalent to: A) 1 N/mm^2 B) 10^6 N/mm^2 C) 10^-3 N/mm^2 D) 1000 N/mm^2