you have a yearly
IN-flux of 25 000 acre feet of water, since i hate working with imperial measurements, this is 3.084 x 10^10 L or
3.084 x 10^7 m³ of water /year
your out-flux is split into two, one being - 0.16ft/day (bottom seep) and the second being -40 inches / year (evaporation)
let's set both to yearly fluxes and SI units, making it
-0.0488m/day -> -17.8m/year (bottom seep)
and
-1.016m/year (evaporation)
both totaling for an out-flux of -18.816m/year of dissipating water ....
so what is asked is a surface area of the basin we're looking at, if we multiply our out-flux velocity with this area we get the total discharge of the basin (this is volume X/year)
in this situation we want this discharge to equal the yearly in-flux of 3.084 x 10^7 m³ /year
so let's call this area or surface (A)
this gives us the equation:
(IN-Flux) + (Out flux) = 0
(3.084 x 10^7 m³ /year) + ( -18.816m/year * (A) ) = 0
=> 18.816m/year * (A) = 3.084 x 10^7 m³ /year
=> (A) = (3.084 x 10^7 m³ /year) / (18.816m/year)
=> (A) = 1.639 x 10^6 m²
or
=> (A) = 405 acres
since we assume a circular reservoir we know that R²*PI = Area
so => R² * PI = 1.639 x 10^6 m²
=> R = +Sqrt (1.639 x 10^6 m² / PI)
R = 722m or 2370 ft or 789.9 yards
