Tag Archives: hydrology

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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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 5

List seven major factors that determine a watershed’s response to a given rainfall.

Solution:

The seven major factors that determine a watershed’s response to a given rainfall are:

  1. Drainage Area
  2. Channel Slope
  3. Soil Types
  4. Land Use
  5. Land Cover
  6. Main Channel and tributary characteristics-channel morphology
  7. The shape, slope, and character of the floodplain
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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 4

Explain how air masses are classified. Where are these types of air masses located?

Solution:

They are classified in two ways: the source from which they are generated, land (continental) or water (maritime), and the latitude of generation (polar or tropical).

These air masses are present in the United States. The Continental polar emanates from Canada and passes over the northern United States. The maritime polar air mass also comes southward from the Atlantic Coast of Canada and affects the New England states. Another maritime polar comes from the Pacific and hits the extreme northwestern states. The maritime tropical air masses come from the Pacific, the Gulf of Mexico, and the Atlantic (these affect the entire Southern United States). Continental tropical air masses form only during summer. They originate in Texas and affect the states bordering the north.

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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 3

Explain the difference between humidity and relative humidity.

Solution:

Humidity is a measure of the amount of water vapor in the atmosphere and can be expressed in several ways. Specific humidity is a mass of water vapor in a unit mass of moist air while relative humidity is a ratio of the air’s actual water vapor content compared to the amount of water vapor at saturation for that temperature.

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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 2

Who is responsible for the first recorded rainfall measurements? Describe the technique used to obtain these measurements.

Solution:

The first recording was obtained in the seventeenth century by Perrault. He obtained his data by comparing measured rainfall to the estimated flow in the Seine River to show how the two were related

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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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Hydrology and Floodplain Analysis, 5th Edition by Philip E. Bedient, Wayne C. Huber, & Baxter E. Vieux Solution Manual

You can purchase the complete solution manual here.

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 13: Case Studies in Hydrologic Engineering: Water Resource Project

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