Question

A drainage basin bordering the sea has an area of 7500 km2 . The average precipitation for this drainage basin is 900 mm year-1 . The average surface water flow at the outlet of the drainage basin equals 22.5 × 108 m3 year-1 . The average groundwater flow to the sea is 100 mm year-1 . The averages are determined for 30 hydrological years. Determine the average actual evaporation in mm year-1 and m3 year-1.

Answer 1

`7.8 * 10^8 m3/y`

`104 mm / yr`

Step-by-step explanation:

Given data:

Basin Area = 7500 km^2

average precipitation
`= 900 mm year^(-1)`

average surface flow
`= 22.5* 10^8 m^3 year^(-1)`

average ground flow
`= 100 mm year^(-1)`

total time period for averages = 30 hydrological year

Volume = height * area ,

here, height of precipitation

total precipitation over the surface
`= (900 * 10^(-3))`

`volume  = (900 * 10^(-3)* (7500 * 10^6) )m3/yr = 37.8* 10^8 m3/yr`

volume of flow of Groundwater to sea

`= ( (100 * 10^(-3)) * (7500* 10^6) )m3/yr = 7.5* 10^8 m3/yr`

Average surface flow
`= 22.5 * 10^8 m3/yr`

The evaporation volume is given as

* Precipitation = Evaporation + surface flow + groundwater flow.

therefore, we have

Evaporation = Precipitation - (surface water flow + groundwater flow)

putting all value in above equation to get evaporation value

Evaporation
`= 37.8 * 10^8 m3/yr - ( 7.5 * 10^8 + 22.5* 10^8 ) m3/yr = 7.8 * 10^8 m3/yr`

obtained height of water by dividing above calculated volume of evaporation by total area

`= (7.8 * 10^8 m^3)/(75 * 10^8 m^2) = 0.104 m = (0.104 * 1000) mm = 104 mm / yr`