Author Archives: Engineering Math

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:

\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 \text{Evaporation}, \ E.

We shall use the formula

\Delta S=I+P-O-E

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

E=I+P-O-\Delta S

Before we can substitute all the given values, we need to convert everything to the same unit of 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*}
\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*}
\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

\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:

\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 S

In 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 S

Before 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*}
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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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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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