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.
a. Draw up the water balance.


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

Answer 1

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

$900=E+\frac{22.5\times 10^{8}}{7500\times 10^{6}}\times 1000+100\\\\900=E+300+100$

Part b)

From the above equation we get

$E=900-400=500mm/year$

The evaporation in cubic meters per year is

$E'=0.5\times 7500\times 10^{6}=3750\times 10^{6}m^3/year$