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Hydrology and Floodplain Analysis Solution Manual

Hydrology and Floodplain Analysis 5th Edition Solution Guide

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Chapter 1: Hydrologic Principles


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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Algebra ENGG10: Age Word Problem


Peter is 36 years old. Peter is twice as old as Jun was when Peter was as old as Jun is now. How old is Jun?


Solution:

Let x be the age of Jun now.

Consider the following table:

[wpdatatable id=1]

Therefore,

\begin{align*}
36-x & =x-\frac{x}{2} \\
36-x & = \frac{x}{2} \\
36 & = \frac{x}{2}+x \\
36 & = \frac{3x}{2} \\
x & = \frac{36\left( 2 \right)}{3} \\
x & = 24 \  \text{years old} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Jun is 24 years old now.


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College Physics by Openstax Chapter 3 Problem 22


A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure 3.60, and then correctly calculates the length and orientation of the fourth side D. What is his result?

A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure 3.60, and then correctly calculates the length and orientation of the fourth side D.
Figure 3.60

Solution:

For the four-sided plot to be closed, the resultant displacement of the four sides should be zero. The sum of the horizontal components should be zero, and the sum of the vertical components should also be equal to zero.

We need to solve for the components of each vector. Take into consideration that rightward and upward components are positive, while the reverse is negative.

For vector A, the components are

\begin{align*}
A_x & = \left( 4.70 \ \text{km} \right) \cos 7.5^\circ \\
A_x & = 4.6598 \ \text{km}
\end{align*}
\begin{align*}
A_y & = -\left( 4.70 \ \text{km} \right) \sin 7.5^\circ \\
A_y & = -0.6135 \ \text{km}
\end{align*}

The components of vector B are

\begin{align*}
B_x & =-\left( 2.48 \ \text{km} \right) \sin 16^\circ \\
B_x & = -0.6836 \ \text{km}

\end{align*}
\begin{align*}
B_y & =\left( 2.48 \ \text{km} \right) \cos 16^\circ \\
B_y & =2.3839 \ \text{km}
\end{align*}

For vector C, the components are

\begin{align*}
C_x & = -\left( 3.02 \ \text{km} \right) \cos 19^\circ \\
C_x & = -2.8555 \ \text{km}
\end{align*}
\begin{align*}
C_y & = \left( 3.02 \ \text{km} \right) \sin 19^\circ \\
C_y & = 0.9832 \ \text{km}
\end{align*}

Now, we need to take the sum of the x-components and equate it to zero. The x-component of D is unknown.

\begin{align*}
A_x+B_x+C_x+D_x & =0 \\
4.6598 \ \text{km}-0.6836 \ \text{km}-2.8555 \ \text{km}+ D_x & =0 \\
1.1207 \ \text{km} +D_x & =0 \\
D_x & = -1.1207 \ \text{km}
\end{align*}

We also need to take the sum of the y-component and equate it to zero to solve for the y-component of D.

\begin{align*}
A_y +B_y+C_y+D_y & =0 \\
-0.6135 \ \text{km}+2.3839 \ \text{km}+0.9832 \ \text{km}+ D_y & =0 \\
2.7536 \ \text{km} +D_y & =0 \\
D_y & = -2.7536 \ \text{km}
\end{align*}

To solve for the distance of D, we shall use the Pythagorean Theorem.

\begin{align*}
D & = \sqrt{\left( D_x \right)^2+\left( D_y \right)^2} \\
D & = \sqrt{\left( -1.1207 \ \text{km} \right)^2+\left( -2.7536 \ \text{km} \right)^2} \\
D & = 2.97 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Then we can solve for θ using the tangent function. Since it is taken from the vertical axis, it can be solved by:

\begin{align*}
\theta & = \tan^{-1} \left| \frac{D_x}{D_y} \right|
 \\
\theta & = \tan^{-1} \left| \frac{-1.1207 \ \text{km}}{-2.7536 \ \text{km}} \right|
\\
\theta & = 22.1 ^ \circ  \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

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College Physics by Openstax Chapter 3 Problem 21


You fly 32.0 km in a straight line in still air in the direction 35.0º south of west. (a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.) (b) Find the distances you would have to fly first in a direction 45.0º south of west and then in a direction 45.0º west of north. These are the components of the displacement along a different set of axes—one rotated 45º.


Solution:

Part A

Consider the illustration shown.

The south and west components of the 32.0 km distance are denoted by DS and DW, respectively. The values of these components are solved below:

\begin{align*}
D_S & = \left( 32.0\ \text{km} \right) \sin 35.0 ^\circ \\
D_S & = 18.4^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}
\begin{align*}
D_W & = \left( 32.0\ \text{km} \right) \cos 35.0 ^\circ \\
D_W & = 26.2^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Part B

Consider the new set of axes (X-Y) as shown below. This new set of axes is rotated 45° from the original axes. Thus, axis X is 45° south of west, and axis Y is 45° west of north. First, we can obviously see that θ has a value of 10°.

Therefore, the components of the 32.0 km distance along X and Y axes are:

\begin{align*}
D_X & = \left( 32.0 \ \text{km} \right) \cos 10^\circ  \\
D_X & = 31.5^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}
\begin{align*}
D_Y & = \left( 32.0 \ \text{km} \right) \sin 10^\circ  \\
D_Y & = 5.56^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

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College Physics by Openstax Chapter 3 Problem 20


A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors A from B in Figure 3.59. She then correctly calculates the length and orientation of the third side C. What is her result?

Figure 3.59

Solution:

Consider the illustration shown.

We need to solve for an interior angle of the triangle. So, we need to solve for the value of α first. This can be done by simply subtracting the sum of 21 and 11 degrees from 90 degrees.

\begin{align*}
\alpha & = 90 ^ \circ -\left( 21^\circ +11^\circ  \right) \\
\alpha & = 58^\circ 
\end{align*}

Then, using the cosine law, we can now solve for the magnitude of vector C. That is

\begin{align*}
C^2 & = A^2 + B^2  - 2AB \cos \alpha \\
C^2 & = \left( 80\ \text{m} \right)^2+\left( 105\ \text{m} \right)^2-2\left( 80\ \text{m} \right)\left( 105\ \text{m} \right) \cos 58^\circ  \\
C^2 & = 8522.3564 \\
C & = \sqrt{8522.3564} \\
C & = 92.3 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Before we can solve for the value of θ, we need to know the value of β first. This can be done by using the sine law.

\begin{align*}
\frac{\sin \beta}{80\ \text{m}} & = \frac{\sin 58^\circ }{92.3 \ \text{m}} \\
\sin \beta & = \frac{\left( 80 \ \text{m} \right)\sin 58^\circ }{92.3 \ \text{m}} \\
\beta & = \arcsin \left[ \frac{\left( 80 \ \text{m} \right)\sin 58^\circ }{92.3 \ \text{m}} \right] \\
\beta & = 47.3^\circ 
\end{align*}

Finally, we can solve for θ.

\begin{align*}
\theta & = \left( 90 ^\circ +11^\circ  \right) - \beta \\
\theta & = \left( 90 ^\circ +11^\circ  \right) - 47.3^\circ  \\
\theta & = 53.7^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

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College Physics by Openstax Chapter 3 Problem 18


You drive 7.50 km in a straight line in a direction 15º east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.) (b) Show that you still arrive at the same point if the east and north legs are reversed in order.


Solution:

Part A

Consider the illustration shown.

Let DE be the east component of the distance, and DN be the north component of the distance.

\begin{align*}
D_E & = 7.50 \  \sin 15^\circ  \\
D_E & = 1.9411 \ \text{km} \\
D_E & = 1.94 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}
\begin{align*}
D_N & = 7.50 \  \cos 15^\circ  \\
D_N & = 7.2444\ \text{km} \\
D_N & = 7.24 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Part B

It can be obviously seen from the figure below that you still arrive at the same point if the east and north legs are reversed in order.


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