Question

using the energy balance method and usual simplifying assumptions, estimate the evaporation rate (mm/d) and yearly evaporation ( m^3/yr) from a lake with a surface area pf 100ha if the net solar radiation is 200 W/m^2

Answer 1

1ha= 10000m2, 100ha= 100*10⁴m2=10⁶ m2 , solar radiation =200W/m2. Assuming 30% of the radiation will be 30% utilized, hence solar radiation used for evaporation = 200*0.3=60 W/m2

Evaporation in terms of energy= 60*10⁶ W =60*10⁶ J/s =60*1000Kj/s , heat of vaporization = 2260 J/gm=2.260Kj/gm, hence water evaporated= 60*1000/(2.260) gm/s =26549 gm/s=26.549 Kg/s

density of water =1000 kg/m3, hence evaporation loss = 26.549/1000 m3/s =0.0265 m3/s

heat of vaporization in terms of Kj/kg =2260 Kj/kg, density =1000 kg/m3, hence heat of vaporization/m3 = 2260Kj/kg*1000kg/m3= 226*10⁴ KJ/m3, evaporatino loss =60*10^-3 Kj/sm²/ (226*10⁴ KJ/m3)=2.65*10-8 m/s

=2.65*10^-8*1000mm/s =2.65*10-5 mm/s = 2.65*10^-5*86400 mm/day=2.29 mm/day

evaporatino loss = 60 W/m2

day =24hrs, 1hr= 60 min, 1 min= 60 sec, 1day= 24*60*60 =86400 s

evaporation loss/day= 0.0265*86400 m3/day =2289.6 m3/day

365 days= 1 year, evaporation loss = 2289.6*365 m3/year =835704 m3/year