Problem:
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*}Hydrology and Floodplain Analysis 5th Edition by Bedient et. al. Chapter 1 Problems
Hydrology and Floodplain Analysis 5th Edition by Bedient, Huber and Vieux
Chapter 1: Hydrologic Principles
Chapter 2: Hydrologic Analysis
Chapter 3: Frequency Analysis
Chapter 4: Flood Routing
Chapter 5: Hydrologic Simulation Models
Chapter 6: Urban Hydrology
Chapter 7: Floodplain Hydraulics
Chapter 8: Ground Water Hydrology
Chapter 9: Design Applications in Hydrology
Chapter 10: GIS Applications in Hydrology
Chapter 11: Radar Rainfall Applications in Hydrology
Chapter 12: Severe Storm Impacts and Flood Management
Chapter 9: Design Applications in Hydrology
Chapter 10: GIS Applications in Hydrology
Chapter 11: Radar Rainfall Applications in Hydrology
Chapter 12: Severe Storm Impacts and Flood Management
Chapter 13: Case Studies in Hydrologic Engineering: Water Resource Project
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