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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.

a. Draw up the water balance.


b. Determine the average actual evaporation in mm year-1 and m3 year-1

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Answer with Step-by-step explanation:

The basic water balance dictates that "Amount of water flowing into the basin should be equal to the amount of water flowing out of it".

Mathematically


Water_(in)=Water_(out)\\\\P=E+Q+I

where

P = precipitation

I = Infiltration losses in the basin

E = Evapotranspiration

Q = is outflow from the basin

Applying the given values we get


900=E+(22.5* 10^(8))/(7500* 10^(6))* 1000+100\\\\900=E+300+100

Part b)

From the above equation we get


E=900-400=500mm/year

The evaporation in cubic meters per year is


E'=0.5* 7500* 10^(6)=3750* 10^(6)m^3/year

User Manoel Campos
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