The total observed runoff volume during a 6h storm with a uniform intensity of 1.5cm/hr is 23.04×106 m³ from a basin of 320 km² of area. The average infiltration rate for the basin is........ mm/hr
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- Average runoff from a 600 km2 catchment area was measured as 60 m3/s 3 points for a 3-hr period .If the depression storage is 10 cm and infiltration is 0.8 in what is the depth of rainfall (precipitation) in m ?An infiltration capacity curve prepared for a catchment indicated an initial infiltration capacity of 2.5 cm/hr and attains a constant value of 0.5 cm/hr in 10 hrs of rainfall with a Horton’s constant k= 6/day. Determine the infiltration loss when t1= 5.5 hrs and t2= 6.9 hrsAnswer these both questions a. The total observed runoff volume during a 4 hr storm with uniform intensity of 2.8cm/hr is 25.2 x106 m3 from basin of 280km2 area. What is the average infiltration rate for the area. b. A volume of 3x106 m3 of groundwater was pumped out from unconfined aquifer uniformly over an area of 5 km2. The pumping lowered the water table from 102 to 99m. The specific yield of the aquifer is?
- A reservoir is in a region where the average annual precipitation is 3106 mm and normal annual US class A pan evaporation is 3456 mm. The average area of reservoir water surface is 50 km2. If under conditions of 30% of the rainfall on the land occupied by the reservoir runoff into the stream, estimate the net annual increase or decrease in the stream flow. Assume suitable pan evaporation coefficient.?If the runoff from an area of 22 m^2 is 16.5 L/min, determine the infiltration capacity under a constant rainfall intensity of 90 mm/hr.A reservoir has an average area of 50 km over an year. The normal annual rainfall at the place is 120 cm and the class A pan evaporation is 240 cm. Assuming the land flooded by the reservoir has a runoff coefficient of 0.45. Determine the runoff volume in Mm^3.
- A catchment area 5000 km2. In one day, the depth of precipitation is 55mm, Evapotranspirationis 0.8 mm/day and the increase in the water in the catchment is 0.3*109 m3/day and the infiltrationrate in this day is 0. 3*109 m3/day.It is required to calculate:(a) Volume of Runoff(b) Average discharge in the river(c) Coefficient of Runoff and what is the type of soil approximately.For a small catchment, the infiltration rate at the beginning of rain was observed to be 90 mm/hr and decreased exponentially to a constant rate of 8 mm/hr after 2.5 hr was 50 mm. Develop the Horton's equation for the infiltration rate at any time t < 2.5 hr.A 1.5 hr storm of intensity 25 mm/hr occurs over a basin for which the Horton’s equation is established as f = 6 + 16 e –2t. Determine the depth of infiltration in the first 45 min and the average infiltration rate for the first 75 min
- For a catchment in India, the mean monthly temperatures are given. Estimate the runoff and annual runoff coefficient by Khosla’s method.A reservoir has an average area of 50 km2 over a year. The normal annual rainfall at the place is 120 cm and the class A pan evaporation is 240 cm. Assuming the land flooded by the reservoir has a runoff coefficient 0.4, estimate the net annual increase or decrease in the stream flow as a result of the reservoir.For the following rainfall rate, find the net runoff in cm, total rainfall and the value of W-index. Take Ø-index = 3.