Clear Lake has a surface area of 708,000 m2 (70.8 ha.). For a given month, the lake has an inflow of 1.5 m3/s and an outflow of 1.25 m3/s. A +1.0-m storage change or increase in lake level was recorded. If a precipitation gage recorded a total of 24 cm for this month, determine the evaporation loss (in cm) for the lake. Assume that seepage loss is negligible.
Solution:
We are given the following values:
\begin{align*}
\text{Area}, \ A&=708,000 \ \text{m}^2 \\
\text{Inflow}, \ I&=1.5 \ \text{m}^3/\text{s} \\
\text{Outflow}, \ O & = 1.25 \ \text{m}^3/\text{s} \\
\text{change in storage}, \ \Delta S & = 1.0 \ \text{m} \\
\text{Precipitation}, \ P&=24 \ \text{cm} \\
\text{time}, \ t &= 1 \ \text{month} = 30 \ \text{days}
\end{align*}The required value is the \text{Evaporation}, \ E.
We shall use the formula
\Delta S=I+P-O-E
Solving for E in terms of the other variables, we have
E=I+P-O-\Delta S
Before we can substitute all the given values, we need to convert everything to the same unit of cm.
\begin{align*}
\text{Inflow}&=\frac{1.5\:\frac{\text{m}^3}{\text{s}}\cdot \frac{100\:\text{cm}}{1\:\text{m}}\cdot \frac{3600\:\text{s}}{1\:\text{hr}}\cdot \frac{24\:\text{hr}}{1\:\text{day}}\cdot \frac{30\:\text{days}}{1\:\text{month}}\cdot 1\:\text{month}}{708,000\:\text{m}^2} \\
\text{Inflow}&=549.1525 \ \text{cm}
\end{align*}\begin{align*}
\text{Outflow}&=\frac{1.25\:\frac{\text{m}^3}{\text{s}}\cdot \frac{100\:\text{cm}}{1\:\text{m}}\cdot \frac{3600\:\text{s}}{1\:\text{hr}}\cdot \frac{24\:\text{hr}}{1\:\text{day}}\cdot \frac{30\:\text{days}}{1\:\text{month}}\cdot 1\:\text{month}}{708,000\:\text{m}^2} \\
\text{Outflow}&=457.6271 \ \text{cm}
\end{align*}\Delta S=1.0 \ \text{m} \times \frac{100 \ \text{cm}}{1.0 \ \text{m}}=100 \ \text{cm}Now, we can substitute the given values in the formula
\begin{align*}
E & =I+P-O-\Delta S \\
E& =549.1525 \ \text{cm}+24 \text{cm}-457.6271 \ \text{cm}-100 \ \text{cm} \\
E& =15.5254 \ \text{cm} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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