Either solve whole question or don't solve at all... Vll upvote A small drainage basin has an area of 5.1 km² and received 380 mm of precipitation in a given year. In the same year, thestream draining this basin had an average discharge of 1.0 x 106 m³ /yr. Assume that there has been no change in waterstored within the basin. Compute the annual evaporation from the basin in m³ /yr. Convert everything to standard unit:x108 m²1. Basin area =2. Precipitation =m/yr3. Annual precipitationamount converted to m³/yr is4. Annual discharge isx 106 m³ /yr.5. Annual storage change AS = 0 (no change)Water balance equation can be written as P = E + Q + AS/Atx106 m³/yr. 1. Plug the value into the equation, the annual evaporation from the basin is2. Convert this to the average evaporation rate in the basin in mm/yr isx106m³/yrmm/yr

1. A small drainage basin has an area = 5.1 km^2We know 1 km^2 = 10^6 sq. mThus the basin area =…

1. Basin area = 2. Precipitation = 3. Annual precipitation amount converted to m³/yr is 4. Annual discharge is x 108 m³ /yr. 5. Annual storage change AS = 0 (no change) -Water balance equation can be written as P = E + Q + AS/At x106 m² m/yr x108 m³/yr -1. Plug the value into the equation, the annual evaporation from the basin is 2. Convert this to the average evaporation rate in the basin in mm/yr is x106m³/yr mm/yr

1. Basin area = 2. Precipitation = 3. Annual precipitation amount converted to m³/yr is 4. Annual discharge is x 108 m³ /yr. 5. Annual storage change AS = 0 (no change) -Water balance equation can be written as P = E + Q + AS/At x106 m² m/yr x108 m³/yr -1. Plug the value into the equation, the annual evaporation from the basin is 2. Convert this to the average evaporation rate in the basin in mm/yr is x106m³/yr mm/yr