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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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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 6

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:

Area, A=1050 acresTime, t=1 monthInflow, I=33 cfsOutflow, O=27 cfsGround seepage, G=1.5 inPrecipitation, P=4.5 inEvaporation, E=6.0 in\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:

∑Inflows−∑Outflows=Change in Storage,ΔS∑I−∑Q=ΔS\sum \text{Inflows}-\sum \text{Outflows}=\text{Change in Storage}, \Delta S \\ \\ \sum I-\sum Q=\Delta S

In this case, the inflows are Inflow I\text{Inflow} \ I and Precipitation, P\text{Precipitation}, \ P, while the others are outflows. Our formula now becomes

∑Inflow−∑Outflow=ΔS(I+P)−(O+G+E)=ΔS\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 S

Before substituting, we need to convert all the given to inches. More specifically the outflow OO and inflow II.

The inflow and outflow, in cfs, will be divided by the given are to come up with units of inches.

The inflow is

Inflow=33 ft3s⋅1 acre43560 ft2⋅12 in1 ft⋅3600 s1 hr⋅24 hr1 day⋅30 days1 month⋅1 month1050 acresInflow=22.4416 in\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

Outflow=27 ft3s⋅1 acre43560 ft2⋅12 in1 ft⋅3600 s1 hr⋅24 hr1 day⋅30 days1 month⋅1 month1050 acresOutflow=18.3613 in\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

ΔS=(I+P)−(O+G+E)ΔS=(22.4416 in+4.5 in)−(18.3613 in+1.5 in+6 in)ΔS=1.0803 in  (Answer)\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

ΔS in volume=1.0803 in×1050 acres×1 ft12 inΔS in volume=94.5263  (Answer)\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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Chapter 1: Hydrologic Principles

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

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