A lake with a surface area of 1050 acres was monitored over a period of time. During a one-month period, the inflow was 33 cfs, the outflow was 27 cfs, and a 1.5-in. seepage loss was measured. During the same month, the total precipitation was 4.5 in. Evaporation loss was estimated as 6.0 in. Estimate the storage change for this lake during the month.
Solution:
We are given the following values:
\begin{align*}
\text{Area}, \ A&=1050 \ \text{acres} \\
\text{Time}, \ t&=1 \ \text{month} \\
\text{Inflow}, \ I&=33 \ \text{cfs} \\
\text{Outflow}, \ O&=27 \ \text{cfs} \\
\text{Ground seepage}, \ G&=1.5 \ \text{in} \\
\text{Precipitation}, \ P&=4.5 \ \text{in} \\
\text{Evaporation}, \ E&=6.0 \ \text{in}
\end{align*}The formula that we are going to use is:
\sum \text{Inflows}-\sum \text{Outflows}=\text{Change in Storage}, \Delta S \\
\\
\sum I-\sum Q=\Delta SIn this case, the inflows are \text{Inflow} \ I and \text{Precipitation}, \ P, while the others are outflows. Our formula now becomes
\color{Blue} \sum \text{Inflow}-\color{Red} \sum \text{Outflow}=\color{Green} \Delta S
\\
\color{Blue}(I+P)- \color{Red}(O+G+E)=\color{Green}\Delta SBefore substituting, we need to convert all the given to inches. More specifically the outflow O and inflow I.
The inflow and outflow, in cfs, will be divided by the given are to come up with units of inches.
The inflow is
\begin{align*}
\text{Inflow}&=\frac{33\:\frac{\text{ft}^3}{\text{s}}\cdot \frac{1\:\text{acre}}{43560\:\text{ft}^2}\cdot \frac{12\:\text{in}}{1\:\text{ft}}\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}}{1050\:\text{acres}} \\
\text{Inflow}& =22.4416 \ \text{in}
\end{align*}The outflow is
\begin{align*}
\text{Outflow}&=\frac{27\:\frac{\text{ft}^3}{\text{s}}\cdot \frac{1\:\text{acre}}{43560\:\text{ft}^2}\cdot \frac{12\:\text{in}}{1\:\text{ft}}\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}}{1050\:\text{acres}} \\
\text{Outflow}& =18.3613\ \text{in}
\end{align*}Now that everything is in inches, we can now substitute the values in the formula
\begin{align*}
\Delta S&=\left( I+P \right)-\left( O+G+E \right) \\
\Delta S&=\left( 22.4416 \ \text{in}+4.5 \ \text{in} \right)-\left( 18.3613 \ \text{in}+1.5 \ \text{in}+6 \ \text{in} \right) \\
\Delta S&=1.0803 \ \text{in} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}We can also state the change in storage in terms of volume by multiplying the given area
\begin{align*}
\Delta S \ \text{in volume} & =1.0803 \ \text{in}\times 1050 \ \text{acres}\times \frac{1 \ \text{ft}}{12 \ \text{in}}\\
\Delta S \ \text{in volume} & = 94.5263 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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