Answer :
Answer:
[tex]7.8 \times 10^8 m3/y[/tex]
[tex]104 mm / yr[/tex]
Explanation:
Given data:
Basin Area = 7500 km^2
average precipitation [tex]= 900 mm year^{-1}[/tex]
average surface flow [tex]= 22.5\times 10^8 m^3 year^{-1}[/tex]
average ground flow[tex] = 100 mm year^{-1}[/tex]
total time period for averages = 30 hydrological year
Volume = height * area ,
here, height of precipitation
total precipitation over the surface [tex]= (900 * 10^{-3}) [/tex]
[tex]volume = (900 * 10^{-3}\times (7500 * 10^6) )m3/yr = 37.8\times 10^8 m3/yr[/tex]
volume of flow of Groundwater to sea
[tex] = ( (100 * 10^{-3}) \times (7500\times 10^6) )m3/yr = 7.5\times 10^8 m3/yr[/tex]
Average surface flow [tex]= 22.5 \times 10^8 m3/yr[/tex]
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 [tex]= 37.8 \times 10^8 m3/yr - ( 7.5 \times 10^8 + 22.5\times 10^8 ) m3/yr = 7.8 \times 10^8 m3/yr[/tex]
obtained height of water by dividing above calculated volume of evaporation by total area
[tex]= \frac{7.8 * 10^8 m^3}{75 * 10^8 m^2} = 0.104 m = (0.104 \times 1000) mm = 104 mm / yr[/tex]