AERE_160_Midterm_Fall_23
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Iowa State University *
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Course
160
Subject
Aerospace Engineering
Date
Dec 6, 2023
Type
Pages
22
Uploaded by DoctorSkunk3684
AER E 160Midterm Fall 2023Due: October 14
th
11:59pm
Please do the following problems. Show all your work. Most of the points for this
assignment are for the problem-solving process. Work can be either handwritten
or typed out, but all must be neatly done. Points deducted from difficult to read
work will not be given back.
Submit all your work as a single PDF on Canvas. The submission link on Canvas will
only accept PDFs. You will not be able to upload non-PDF files in the submission
link.
This exam is open notes, book, and internet. Do not discuss your work with your
classmates.
You will not be able to get an extension on this assignment. Once the link closes
on Canvas you will have no opportunity to submit the assignment for grading.
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Problem 1 (20 pts)
A balloon in a university building at sea-level has a volume of
. It is filled with perfectly safe
46 ??
3
hydrogen.
a.)
In equilibrium, what is the mass of the payload? Use English units. (5 pts)
Volume (V) = 46 ft^3
Density of Hydrogen (
⍴
) = 0.0051 lb/ft^3
Mass (m) = ?
m = V *
⍴
m = 46 ft^3 * 0.0051 lb/ft^3
m = 0.23 lbs
Therefore, the mass of the payload is equal to 0.23 lbs.
b.)
In equilibrium, what is the
weight
of the payload? Use English units. (5 pts)
Acceleration due to gravity (g) = 32.2 ft/s^2
Weight (W) = ?
W = m * g
W = 0.23 lbs * 32.2 ft/s^2
W = 7.4 lbs
Therefore, The weight of the payload is equal to 7.4 lbs.
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Problem 1 continued:
c.)
If the balloon was filled with helium, what is the
weight
of the payload in equilibrium? Use SI
units. (5 pts)
Volume = 46 ft^3 = 1.30 m^3
Density of Helium (
⍴
) = 0.18 kg/m^3
Acceleration due to gravity (g) = 9.8 m/s^2
Mass (m) = ?
Weight (W) = ?
m =
⍴
* V
m = 0.18 kg/m^3 * 1.30 m^3
m = 0.23 kg
W = m * g
W = 0.23 kg * 9.8 m/s^2
W = 2.25 kg
Therefore, the weight of the payload with Helium is equal to 2.25 kg
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AER E 160Midterm Fall 2023Due: October 14
th
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d.)
What volume of helium would be required to lift a 10 kg payload at sea-level? Use SI units (5 pt)
Mass (m) = 10 kg
Density (
⍴
) = 0.18 kg/m^3
Volume (V) = ?
V = m /
⍴
V = 10 kg / 0.18 kg/m^3
V = 55.5 m^3
Therefore, the volume required to lift a 10 kg payload is equal to 55.5 m^3.
Problem 2 (20 pts) – Standard Atmosphere
a.)
What is the difference between geometric, absolute, and geopotential altitudes? (6 pts)
Geometric altitude, also referred to as geometric height, represents the actual straight-line
distance above a reference point, often the Earth's surface or a reference ellipsoid. This
measurement does not factor in changes in gravitational force or the Earth's shape.
Absolute altitude, on the other hand, denotes the height above the Earth's surface measured
along a line perpendicular to the reference ellipsoid. It does not consider variations in gravity's
strength with altitude, making it less suitable for certain aerodynamic and atmospheric
calculations.
Geopotential altitude takes into account the gravitational potential energy per unit mass and
acknowledges variations in gravitational force with altitude. It is the preferred choice for
atmospheric calculations, providing a consistent measure of altitude even in regions with
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significant gravitational field variations. This measurement is typically expressed in units of
length, such as meters or feet.
b.)
What is the difference between isothermal layers and gradient regions in Standard
Atmosphere. (4pts)
An isothermal layer in the atmosphere refers to a section where the temperature remains
consistent as you go higher in altitude. In the Standard Atmosphere, there are areas where
temperature doesn't alter as you ascend. These layers are primarily located in the lower
atmosphere (troposphere) and the thermosphere.
In gradient regions of the standard atmosphere, temperature changes with altitude, and this
change is typically described using a temperature lapse rate. In these zones, temperature
typically decreases as you move higher in the atmosphere, which is a common occurrence. The
standard atmosphere model assumes a specific temperature lapse rate for these regions.
c.)
What is the value of the acceleration due to gravity (g) at 90,000ft? (5pts)
Height (h) = 90,000 ft = 27.4 km
Radius of Earth (R) = 6378.1 km
Acceleration due to Gravity at Earth’s surface (g
0
) = 9.8 m/s^2
Acceleration due to Gravity at a height (g(h)) = ?
g(h) = g
0
(R / (R + h))^2
g(27.4) = 9.8 (6378.1 / (6378.1 + 27.4))^2
g (27.4) = 9.72 m/s^2 or 31.9 ft/s^2
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Therefore, the acceleration due to gravity at 90,000 ft is equal to 31.9 ft/s^2
d.)
Why do we use geopotential altitude instead of geometric altitude in Standard Atmosphere
calculations? (5pts)
Geopotential altitude is favored over geometric altitude because it offers a more reliable way to
measure altitude in areas where gravity's strength varies more. While geometric altitude is
simple, it doesn't consider the shifting gravitational force at different altitudes. Using geometric
altitude in regions with significant gravity variations, like near the Earth's surface compared to
higher altitudes, would cause inaccuracies. Therefore, geopotential altitude is the preferred
measure for such calculations due to its consistency and accuracy in accounting for these
variations in gravity.
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th
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Problem 3 (20 pts)
Answer the following questions about programming
a.)
What is the difference between a float and an integer? (5 pts)
A float, short for a "floating-point number," is a real number that can represent both whole and
fractional numbers. Floats can have decimal points and can represent a wide range of values,
including very large and very small numbers. They are commonly used in situations that require
precision with decimal value, such as measurements and scientific calculations.
On the other hand, an integer is a whole number that can be either positive, negative, or zero. It
does not contain any fractional or decimal parts, making it suitable for representing whole
quantities, such as counts, indices, or years.
b.)
If you type the command below into your favorite Python 3 IDE (i.e. Spyder, Jupyter
Notebook, etc.) you will get an error. Identify what is incorrect with the command and write
your correction to it so it will successfully run. (5 pts)
3var = [1, ‘cat’, 7.9]
In python, variable names cannot start with a number, so the ‘3’ in the front needs to be
removed. Also, lists cannot contain a mix of data types like integers, strings, and floating-point
numbers in the same list without being enclosed in another data structure, so the elements
should be enclosed in a list or another suitable data structure. The modified command is as
follows:
my_list = [1, 'cat', 7.9]
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Problem 3 continued:
Use the image below to answer c and d
c.)
Line 6 uses np.linspace to generate an array of numbers between 0 and 1. What would the
command look like with np.arange? Feel free to have any number of points between 0 and 1,
but it must start at 0 and end at 1. (5 pts)
When np.arange is used, the command looks as follows:
x = np.arange(0, 1.01, 1 / num_points)
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Problem 4 (40 pts)
A model wing of constant chord length is placed in a low-speed subsonic wind tunnel, spanning the test
section. The wing has a NACA 2415 airfoil and a chord length 1.25m. The flow in the test section is at a
velocity of
at standard sea-level conditions. If the wing is at a
angle of attack calculate the
60
?
?
6°
following with the Appendix D information.
a.)
Reynold’s Number of the airfoil, use
for viscosity at sea level (5pts)
1. 789∙10
−5
??
?∙?
Velocity (v) = 60 m/s
Chord length (l) = 1.25 m
Viscosity (μ) = 1.789 * 10 ^(-5) kg/(m*s)
Air density (
⍴
) = 1.225 kg/m^3
Reynold’s Number (Re) = ?
Re = (
⍴
* v * l) / μ
Re = (1.225 kg/m^3 * 60 m/s * 1.25 m) / (1.789 * 10^(-5) kg/(m*s))
Re = 5.136 * 10^6
Therefore, the Reynold’s Number of the airfoil is equal to 5.136 * 10^6.
b.)
Find
using Appendix D (5pts)
?
?
Angle of attack = 6°
Based on the angle of attack and the graph in appendix D, the Lift Coefficient (
) is equal to 0.8.
?
?
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th
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c.)
Find
using Appendix D (5pts)
?
?
Lift Coefficient (
)
= 0.8
?
?
Based on the Lift Coefficient (
) and the second graph in Appendix D, the Drag Coef±icient (
?
?
) is equal to 0.008.
?
?
d.)
Find the dynamic pressure of the flow (5pts)
Air density (
⍴
) = 1.225 kg/m^3
Velocity (v) = 60 m/s
Dynamic pressure (q) = ?
q = (
⍴
* v^2) / 2
q = (1.225 kg/m^3 * (60 m/s)^2) / 2
q = 2205 Pa
Therefore, the dynamic pressure of the flow is equal to 2205 N/m^2
e.)
Calculate the lift per unit span of the wing (5pts)
Dynamic pressure (q) = 2205 N/m^2
Lift Coefficient (
) = 0.8
?
?
Cord length (l) = 1.25 m
Lift (L) = ?
L = q *
* l
?
?
L = 2205 N/m^2 * 0.8 * 1.25 m
L = 2205 N
Therefore, the Lift per unit span of the wing is equal to 2205 N.
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f.)
Calculate the drag per unit span of the wing (5pts)
Dynamic pressure (q) = 2205 N/m^2
Drag Coefficient (
) = 0.008
?
?
Cord length (l) = 1.25 m
Drag (D) = ?
D = q *
* l
?
?
D = 2205 N/m^2 * 0.008 * 1.25 m
D = 22.05 N
Therefore, the Drag per unit span of the wing is equal to 22.05 N.
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Appendix A
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Appendix B
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Appendix D
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