Clear Lake has a surface area of 70 ha. In April the inflow of the lake was 1.5 m3/sec. A dam regulated the outflow of the lake to be 1.25 m3/sec. If the precipitation recorded for the month of April was 7.62 cm and the storage volume increased by an estimated 650,000 m3. What is the estimated evaporation in cubic meters?

Let suppose that precipitation is very small in comparison with depth of the Clear Lake. The monthly change in the volume of the lake ([tex]\Delta V[/tex]), in cubic meters, is estimated by the following formula:

[tex]\Delta V = A\cdot \Delta z + (\dot V_{in}-\dot V_{out})\cdot \Delta t -V_{evap}[/tex] (1)

Where:

[tex]A[/tex] - Surface area of the lake, in square meters.

[tex]\Delta z[/tex] - Water precipitation, in meters.

[tex]\dot V_{in}[/tex] - Average water inflow, in cubic meters per second.

[tex]\dot V_{out}[/tex] - Average water outflow, in cubic meters per second.

[tex]\Delta t[/tex] - Monthly time, in seconds.

[tex]V_{evap}[/tex] - Evaporation, in cubic meters.

If we know that [tex]A = 700000\,m^{2}[/tex], [tex]\Delta z = 0.0762\,m[/tex], [tex]\dot V_{in} = 1.5\,\frac{m^{3}}{s}[/tex], [tex]\dot V_{out} = 1.25\,\frac{m^{3}}{s}[/tex], [tex]\Delta t = 2.592\times 10^{6}\,s[/tex] and [tex]\Delta V = 650000\,m^{3}[/tex], then the estimated evaporation is:

[tex]650000 = 701340-V_{evap}[/tex] (2)

[tex]V_{evap} = 51340\,m^{3}[/tex]

The estimated evaporation is 51340 cubic meters.