MATLAB for Engineers, 5th Edition Solution Manual

Name: MATLAB for Engineers, 5th Edition Solution Manual
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Author: Zoe Jordan

Unlock the answers to every textbook problem with MATLAB for Engineers, 5th Edition Solution Manual, making studying more efficient and effective.

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Example 1.1
The text book version of this problem was performed in the command window. This version of the
problem is solved using a Live Script
E = 385e24 %Define the rate of energy consumption in J/sec
E = 3.85e+26
E = E*366*24 %Find the energy consumed in one day, in J
E = 3.3818e+30
c = 3.0e8 % Define the speed of light in m/s
c = 300000000
m = E/c^2 % Calculate the mass consumed using Einstein's equation
m = 3.7576e+13

EXAMPLE 2.1
SCALAR OPERATIONS
Wind tunnels (see Figure 2.6) play an important role in our study of the behavior of high-performance
aircraft. In order to interpret wind tunnel data, engineers need to understand how gases behave. The
basic equation describing the properties of gases is the ideal gas law, a relationship studied in detail in
freshman chemistry classes. The law states that
PV = nRT
where P = pressure in kPa,
V = volume in m3,
n = number of kmoles of gas in the sample,
R = ideal gas constant, 8.314 kPa m3/kmol K, and
T = temperature, expressed in kelvins (K).
In addition, we know that the number of kmoles of gas is equal to the mass of the gas divided by the
molar mass (also known as the molecular weight) or
n = m/MW
where
m = mass in kg and
MW= molar mass in kg/kmol.
Different units can be used in the equations if the value of R is changed accordingly.
Figure 2.6 Wind tunnels are used to test aircraft designs. (Louis Bencze/Getty Images Inc., Stone
Allstock.)
Now suppose you know that the volume of air in the wind tunnel is 1000 m3. Before the wind tunnel is
turned on, the temperature of the air is 300 K, and the pressure is 100 kPa. The average molar mass
(molecular weight) of air is approximately 29 kg/kmol. Find the mass of the air in the wind tunnel.
To solve this problem, use the following problem-solving methodology:
1. State the Problem
When you solve a problem, it is a good idea to restate it in your own words: Find the mass of air in a
wind tunnel.
2. Describe the Input and Output
Input
Volume V = 1000 m3
Temperature T = 300 K
Pressure P = 100 kPa
Molecular weight MW = 29 kg/kmol

Gas constant R = 8.314 kPa m3/kmol K
Output
Mass m = ? kg
3. Develop a Hand Example
Working the problem by hand (or with a calculator) allows you to outline an algorithm, which you can
translate to MATLAB® code later. You should choose simple data that make it easy to check your work.
In this problem, we know two equations relating the data:
PV = nRT ideal gas law
n= m/MW relationship between mass and moles
Solve the ideal gas law for n, and plug in the given values:
n= PV/RT
= 40.0930 kmol
Convert moles to mass by solving the conversion equation for the mass m and plugging in the values:
m = n × MW = 40.0930 kmol × 29 kg/mol
m = 1162.70 kg
4. Develop a MATLAB® Solution
First, clear the screen and memory:
clear, clc
Now perform the following calculations in the command window:
P = 100
P = 100
T = 300
T = 300
V = 1000
V = 1000
MW = 29
MW = 29
R = 8.314

R = 8.314
n = (P*V)/(R*T)
n = 40.093
m = n*MW
m = 1162.7
There are several things you should notice about this MATLAB® solution. First, because no semicolons
were used to suppress the output, the values of the variables are repeated after each assignment
statement. Notice also the use of parentheses in the calculation of n. They are necessary in the
denominator, but not in the numerator. However, using parentheses in both makes the code easier to
read.
5. Test the Solution
In this case, comparing the result with that obtained by hand is sufficient. More complicated problems
solved in MATLAB® should use a variety of input data, to confirm that your solution works in a variety of
cases. The MATLAB® screen used to solve this problem is shown in Figure 2.7.

Figure 2.7 MATLAB®screen used to solve the ideal gas problem.
Notice that the variables defined in the command window are listed in the workspace window. Notice
also that the command history lists the commands executed in the command window. If you were to
scroll up in the command history window, you would see commands from previous MATLAB® sessions.
All of these commands are available for you to move to the command window.

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Example 2.3
Calculating Drag
clear,clc
% Find the drag coefficient
drag = 20000;
density = 0.000001;
velocity = 100*0.4470;
area = 1;
cd = drag*2/(density*velocity^2*area)
cd = 2.0019e+07
Find drag for a range of velocities
velocity = 0:20:200;
velocity = 100*velocity;
drag = cd*density*velocity.^2*area/2;
my_table = [velocity',drag']
my_table =
1.0e+09 *
0 0
0.0000 0.0400
0.0000 0.1602
0.0000 0.3603
0.0000 0.6406
0.0000 1.0010
0.0000 1.4414
0.0000 1.9619
0.0000 2.5624
0.0000 3.2431

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Example 3.1
Using the Clausius Clapeyron Equation, find the saturation vapor pressure for water at different
temperatures
%
clear,clc
TempF = [-60:10:120]; %Define temp matrix in F
TempK = (TempF + 459.6)/1.8; %Convert temp to K
Delta_H = 2.45e6; %Define latent heat of vaporization
R_air = 461; %Define ideal gas constant for air
%
Calculate the Vapor Pressures
Vapor_Pressure = 6.11*exp((Delta_H/R_air)*(1/273 - 1./TempK));
Display the results in a table
my_results = [TempF',Vapor_Pressure']
my_results =
-60 0.06979
-50 0.12521
-40 0.21845
-30 0.37138
-20 0.61633
-10 1
0 1.5888
10 2.4749
20 3.7847
30 5.688
40 8.4102
50 12.246
60 17.575
70 24.881
80 34.773
90 48.01
100 65.526
110 88.461
120 118.19

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Example 3_2
clear, clc
% Define the input
Force =[100, 200, 50];
theta = [-90, +90, +30];
%convert angles to radians
theta = theta*pi/180;
%Find the x components
ForceX = Force.*cos(theta);
%Sum the x components
ForceX_total = sum(ForceX);
%Find and sum the y components in the same step
ForceY_total = sum(Force.*sin(theta));
%Find the resulting angle in radians
result_angle = atan(ForceY_total/ForceX_total);
%Find the resulting angle in degrees
result_degrees = result_angle*180/pi
result_degrees = 70.893
%Find the magnitude of the resulting force
Force_total = ForceX_total/cos(result_angle)
Force_total = 132.29

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Example 3.3 - Weather Data
In this example we will find the total precipitation
for each month, and for the entire year, using a data file
We will also find the month and day on which the precipitation was
the maximum
clc
load weatherdata
%Find the sum of each row, which is the sum for each month
monthly_total=sum(weatherdata,2,'omitnan')/100;
monthly_total'
ans = 8.48 5.15 3.85 4.74 3.21 6.33
3.17 2.78 2.49 5.66 4.68 2.86
%Find the annual total
yearly_total = sum(monthly_total)
yearly_total = 53.4
%Find the annual maximum, and the day on which it occurs
[max_precip,month]=max(max(weatherdata,[],2,'omitnan'))
max_precip = 272
month = 1
%Find the annual maximum, and the month in which it occurs
[max_precip,day]=max(max(weatherdata,[],1,'omitnan'))
max_precip = 272
day = 3

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Example 3.4 Climatological Data
In this example we find the mean daily precipitation for each month and the
mean daily precipitation for the year. We also find the standard deviation
of the data
clear, clc
% Changing the format to shortg, often makes the output
% easier to read
format shortg
% By saving the variable weatherdata from the last example, it is
% available to use in this problem
load weatherdata
Average_daily_precip_monthly = mean(weatherdata','omitnan')
Average_daily_precip_monthly = 27.355 18.393 12.419 15.8 10.355
21.1 10.226 8.9677 8.3 18.258
15.6 9.2258
Average_daily_precip_yearly = mean(weatherdata(:),'omitnan')
Average_daily_precip_yearly = 14.63
Now calculate the standard deviation
Monthly_Stdeviation = std(weatherdata','omitnan')
Monthly_Stdeviation = 63.779 36.499 20.401 49.73 26.648 51.176
30.626 30.77 27.447 42.079 54.174 21.012
Yearly_Stdeviation = std(weatherdata(:),'omitnan')
Yearly_Stdeviation = 39.948

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Example 3.5
Noise
load handel %Load the music data file
sound(y,Fs) %Play the music data file
pause %Pause to listen to the music
% Be sure to hit enter to continue after playing the music
% Add random noise
noise=randn(length(y),1)*0.1;
sound(y+noise,Fs)
% Plot the first 200 data points in each file
t=1:length(y);
noisy=y+noise;
plot(t(1,1:200),y(1:200,1),t(1,1:200),y(1:200,1)+noisy(1:200,1),':')
title('Handel''s Messiah')
xlabel('Element Number in Music Array')
ylabel('Frequency')

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Example 4.1
In this example we extract data from a large matrix, and use the data
analysis functions to find the mean high and mean low temperatures for the
year, and to find the high temperature and the low temperature for the year
clear, clc
load the data matrix from a file
load asheville_1999
extract the mean high temperatures from the large matrix
mean_max = asheville_1999(1:12,2);
% extract the mean low temperatures from the large matrix
mean_min = asheville_1999(1:12,3);
% Calculate the annual means
annual_mean_max = mean(mean_max)
annual_mean_max = 68.05
annual_mean_min = mean(mean_min)
annual_mean_min = 46.325
% extract the high and low temperatures from the large matrix
high_temp = asheville_1999(1:12,8);
low_temp = asheville_1999(1:12,10);
% Find the max and min temperature for the year
max_high = max(high_temp)
max_high = 96
min_low = min(low_temp)
min_low = 9
% Create an new matrix with just the temperature information
new_table =[mean_max, mean_min, high_temp, low_temp]
new_table =
51.4 31.5 78 9
52.6 32.1 66 16
52.7 32.5 76 22
70.1 48.2 83 34
75 51.5 83 40
80.2 60.9 90 50
85.7 64.9 96 56
86.4 63 94 54
79.1 54.6 91 39
67.6 45.5 78 28
62.2 40.7 76 26

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MATLAB for Engineers, 5th Edition Solution Manual

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