Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 7

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:

Area, A=708,000 m2Inflow, I=1.5 m3/sOutflow, O=1.25 m3/schange in storage, ΔS=1.0 mPrecipitation, P=24 cmtime, t=1 month=30 days\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 Evaporation, E\text{Evaporation}, \ E.

We shall use the formula

ΔS=I+P−O−E\Delta S=I+P-O-E

Solving for EE in terms of the other variables, we have

E=I+P−O−ΔSE=I+P-O-\Delta S

Before we can substitute all the given values, we need to convert everything to the same unit of cm.

Inflow=1.5 m3s⋅100 cm1 m⋅3600 s1 hr⋅24 hr1 day⋅30 days1 month⋅1 month708,000 m2Inflow=549.1525 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*}
Outflow=1.25 m3s⋅100 cm1 m⋅3600 s1 hr⋅24 hr1 day⋅30 days1 month⋅1 month708,000 m2Outflow=457.6271 cm\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*}
ΔS=1.0 m×100 cm1.0 m=100 cm\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

E=I+P−O−ΔSE=549.1525 cm+24cm−457.6271 cm−100 cmE=15.5254 cm  (Answer)\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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