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AE 202: Aerospace Flight Mechanics

创建日期：2024-05-14 14:59:10

所属分类：物理Physical

标签： AE 202

Aerospace Flight Mechanics

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AE 202: Aerospace Flight Mechanics

Homework Assignment 4 - Solutions
Spring 2021
Problems
1. [20 points] You are helping to design the landing system for a 1000-kg rover that will explore
the surface of Mars. Your team is considering using only a parachute for descent and landing,
i.e. no rocket-powered descent stage. Assume the parachute is a disk-gap-band supersonic
parachute similar to that used on previous Mars landers. Use the data in Figure 1 and Table 1
to answer the following:
(a) What diameter parachute is required to land the rover on Mars with a touchdown veloc-
ity of 0.5 m/s?
Begin with force balance:
FD =W
1
2
ρV 2CDA = mg
Equation above is rearranged to solve for the area of a parachute. The velocity will be
the terminal velocity, gravity is Mar’s gravity, CD is chosen from the table at subsonic
speed, density is taken as the sea level density.
A =
2mgM
ρV 2terminalCD
A =
2(1000 kg)(3.71 m/s2)
(0.02 kg/m3)(.5 m/s)2(.67)
A = 2.215 x 106 m2
D =
√
4A
pi
D =
√
4 (3.305 x 106 m2)
pi
1
The parachute diameter for a terminal velocity of 0.5 m/s is:
D = 1679.3 m
(b) How many g’s of acceleration would the above parachute initially generate if inflated at
Mach 1.7 at an altitude of 7200 m?
The equation for g-forces is given in the presentation as:
Fg =
a
gE
Recall that the gravity is always referenced back to Earth’s gravity. The acceleration
is found using F = ma, Newton’s Second Law, where F is the force of drag at the
specified conditions.
FD =
1
2
ρV 2CDA
a =
ρV 2CDA
2m
Using the graph, CD is found to be 0.775 at M = 1.7. The velocoty must be found
using the Mach equation, where the temperature, ratio of specific heats, and specific gas
constant for Mars is given.
V =Ma
V =M
√
γMarsRMarsT
V = (1.7)
√
(1.3)(188.9
J
kg −K)(210 K)
V = 386.05 m/s
Finally the density at 7200 m is calculated using the exponential atmosphere model with
the Martian atmosphere constants.
ρ = ρ0e
−h
H
ρ = (0.02 kg/m3)e
−7.2 km
11.1 km
ρ = .010455 kg/m3
Plugging all the above values into the g-force equation:
2
a =
ρV 2CDA
2m
a =
(.010455 kg/m3)(386.05 m/s)2(.775)(3.305 x 106 m2)
2(1000 kg)
a = 1337389.8 m/s2
Fg =
a
gE
Fg =
1995511.9 m/s2
9.81 m/s2
Fg = 136329.2 g
(c) Discuss your results.
This is clearly an absurd amount of g’s. This comes from the very small terminal veloc-
ity that was chosen to be reached at h = 0 in part (a). This resulted in a extremely large
parachute area, which translated to high g forces. This should indicate why the recent
martian langers, with large payloads, chosen to use the sky crane system because they
could not reach a safe terminal velocity with parachutes alone.
Table 1: Mars Properties
Volumetric mean radius 3389.5 km
Surface gravitational acceleration 3.71 m/s2
Surface atmospheric density 0.02 kg/m3
Atmospheric scale height 11.1 km
Average atmospheric temperature 210 K
Specific gas constant 188.92 J/(kg K)
Ratio of specific heats 1.3
3
0.4
0.5
0.5
0.6
0.6
0.7
0.7
0.8
0.8
0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
Dr
ag
C
oe
ffi
ci
et
Mach Number
Figure 1: Disk-gap-band parachute drag coefficient.
4
2. [20 points] Compute the equilibrium altitude of a zero-pressure weather balloon (relative to
standard atmosphere) if the payload mass is 15 kg, the maximum volume of the ballon is
800 m3, and:
(a) the lift gas is helium, RHe ≈ 2077.1 J/(kg K)
(b) the lift gas is hydrogen, RH2 ≈ 4124.2 J/(kg K)
(c) the lift gas is methane, RCH4 ≈ 518.28 J/(kg K)
Start with a balance of forces (buoyancy and gravity),
∑
F = 0
Fb + Fg = 0
ρairVdispg −mg = 0
Then, assume Vdisp = Vgas = V and, for zero-pressure balloon, Pgas = Patm and Tgas =
Tatm, so with ideal gas law P = ρRT , can write
mp =
V Patm
Tatm
(
1
Rair
− 1
Rgas
)
Solve for P/T ratio,
Patm
Tatm
=
mp
V
(
1
Rair
− 1
Rgas
)−1
and combine with ideal gas law,
Patm
Tatm
= ρatmRair
ρatmRair =
mp
V
(
1
Rair
− 1
Rgas
)−1
ρatm =
mp
V
(
1− Rair
Rgas
)−1
then can calculate density for each lift gas
(a) helium:
ρatm =
15 kg
800 m3
(
1−
287.058 Jkg K
2077.1 Jkg K
)−1
ρatm = 0.02176 kg/m
3
5
(b) hydrogen:
ρatm =
15 kg
800 m3
(
1−
287.058 Jkg K
4124.2 Jkg K
)−1
ρatm = 0.02015 kg/m
3
(c) methane:
ρatm =
15 kg
800 m3
(
1−
287.058 Jkg K
518.28 Jkg K
)−1
ρatm = 0.042028 kg/m
3
Finally, can look up densities in standard atmosphere table and match with altitude
(geopotential) to find the equilibrium altitude for the zero-pressure weather balloon.
(a) If the lift gas is helium, the equilibrium altitude is
h(ρ) ≈ 28, 800 m
(b) If the lift gas is hydrogen, the equilibrium altitude is
h(ρ) ≈ 29, 150 m
(c) If the lift gas is methane, the equilibrium altitude is
h(ρ) ≈ 24, 700 m
6
3. [20 points] Consider a super-pressure balloon designed to operate in the Venus atmosphere. If
the required payload mass is 100 kg, 10 kg of lift gas is used, and the balloon envelope has an
areal mass of 0.75 kg/m2, what is the required volume of the balloon if the desired operational
altitude is 60 km? Assume an exponential atmosphere (ρsurface = 65 kg/m3, H = 15.9 km).
Starting with Archimedes’ Principle,
Fb + Fg = 0
ρairVdispg −mg = 0
ρairVdispg −
(
mp +mgas +me
)
g = 0
ρairVdisp −
(
mp +mgas +me
)
= 0
where me is the mass of the envelope of the balloon. Assuming the balloon is a sphere, can
describe mass using areal mass (ρA) and surface area (S),
mb = ρAS
where surface area is
S = 4pir2
Since we want to find required volume, can write mass of envelope as a function of volume,
V =
4
3
pir3
r =
(
3V
4pi
) 1
3
S = 4pi
(
3V
4pi
) 2
3
me = ρA4pi
(
3V
4pi
) 2
3
Going back to the balance of forces of buoyancy and gravity, can write polynomial with
volume as the only unknown. Assume Vdisp = Vgas = V (i.e. neglect thickness of balloon
envelope),
ρairV −mp −mgas − ρA4pi
(
3V
4pi
) 2
3
= 0
7
Can calculate density of the air using exponential atmosphere model,
ρair(h) = ρSLe
(
−h
H
)
ρair(h) = (65 kg/m
3)e
(
−60 km
15.9 km
)
ρair = 1.493 kg/m
3
Now, using MATLAB root solver, can find volume
(
1.493 kg/m3
)
V − (100 kg)− (10 kg)− (0.75 kg/m2)(4pi)(3V
4pi
) 2
3
= 0
∴ the required volume of the balloon is
V = 138.8 m3
8
4. [20 points] A wing with a rectangular planform mounted in a low-speed subsonic wind tunnel.
The wing model completely spans the test section so that it may be considered an infinite
wing. If the wing has a NACA 4412 airfoil section and a chord of 0.4 m, calculate the lift,
drag, and moment about the quarter-chord per unit span when the freestream air pressure
is 98 kPa, the freestream temperature is 280 K, the freestream velocity is 38 m/s, and the
viscosity is 1.7894× 10−5 kg/(m s) at the following angles of attack:
(a) 5 deg
(b) 16 deg
(c) -2 deg
You can find the required data for the NACA 4412 airfoil on http://airfoiltools.com/search.
Type “4412” in the “Text search” box and click “Search.” Click “Airfoil details” for the
NACA 4412 in the search results. Under “Polars for NACA 4412”, click “details” for the
appropriate Reynolds Number (use Ncrit = 9 for wind tunnels). You can then scroll through
the airfoil data or download a text file.
First, calculate freestream density using ideal gas law
P∞ = ρ∞RT∞
ρ∞ =
P∞
RT∞
ρ∞ =
98× 103 Pa(
287.058 Jkg K
)(
280 K
)
ρ∞ = 1.2193 kg/m3
Then find dynamic pressure,
q∞ =
1
2
ρ∞v2∞
q∞ =
1
2
(1.2193 kg/m3)(38 m/s)2
q∞ = 880.31 Pa
Next, calculate Reynolds number,
Re =
ρ∞v∞c
µ∞
Re =
(1.2193 kg/m3)(38 m/s)(0.4 m)
1.7894× 10−5 kg/(m s)
Re = 1, 035, 701.4
9
Use Reynolds number to find required data for the NACA 4412 airfoil,
α,◦ Cl Cd Cm
5 1.0254 0.00797 -0.0998
16 1.6698 0.04827 -0.0447
-2 0.2625 0.00715 -0.1038
Table 2: NACA 4412 airfoil data forRe = 1035701,Ncrit = 9
Using the data in Table 2 , can calculate lift, drag, and moment about the quarter-chord pre
unit span using the following expressions,
L = q∞SrefCl
D = q∞SrefCd
M c
4
= q∞SrefCmc
Since we have an infinite wing,
Sref = c(1)
Sref = (0.4 m)(1 m)
Sref = 0.4 m
2
Sample calculation for α = 5◦,
L = q∞SrefCl
L = (880.31 Pa)(0.4 m2)(1.0254)
L = 361.07 N per unit span
D = q∞SrefCd
D = (880.31 Pa)(0.4 m2)(0.00797)
D = 2.81 N per unit span
M c
4
= q∞SrefCmc
M c
4
= (880.31 Pa)(0.4 m2)(1.0254)(0.4 m)
M c
4
= −14.06 N m per unit span
10
(a) For an angle of attack α = 5◦, the lift, drag, and moment about the quarter-chord
are
L = 361.07 N per unit span
D = 2.81 N per unit span
M c
4
= −14.06 N m per unit span
(b) For an angle of attack α = 16◦, the lift, drag, and moment about the quarter-chord
are
L = 587.98 N per unit span
D = 17.0 N per unit span
M c
4
= −6.30 N m per unit span
(c) For an angle of attack α = −2◦, the lift, drag, and moment about the quarter-chord
are
L = 92.43 N per unit span
D = 2.52 N per unit span
M c
4
= −14.62 N m per unit span
11
5. [20 points] This problem requires the use of 3 data files containing inviscid CP values for a
NACA 2412 airfoil. The data is stored in comma-separated value (CSV) text files.
Filename Angle of attack
naca2412i_1.csv 0 deg
naca2412i_2.csv 5 deg
naca2412i_3.csv -5 deg
(a) For each angle of attack, co-plot CP versus x/c with an inverted vertical axis (i.e. neg-
ative values above horizontal axis) for the upper and lower surfaces.
(b) Write a computer program/script to approximate cl for each angle of attack with CP
data in the files using the method discussed in class. To help you check your program,
cl ≈ 0.256 for a NACA 2412 airfoil at zero angle of attack.
(c) Create a plot of cl as a function of α using the computed cl values. What is the zero-lift
angle of attack for this airfoil?
(d) Discuss your results.
(e) Include your computer program/script as text in your assignment.
(a) Figures 2-4 shown below contain plots for CP vs x/c for the upper and lower surfaces
for various angles of attack.
Figure 2: NACA 2412 airfoil CP vs x/c, α = 0◦
12
Figure 3: NACA 2412 airfoil CP vs x/c, α = 5◦
Figure 4: NACA 2412 airfoil CP vs x/c, α = −5◦
(b) To approximate Cl, need to use the following expression,
Cl =
1
c
∫ c
0
(
CP,l − CP,u
)
dx
13
Finding the area between the data sets in Figures 2-4 results in the following
lift coefficients,
Cl(α = 0
◦) = 0.2564
Cl(α = 5
◦) = 0.8294
Cl(α = −5◦) = −0.3181
(c) These lift coefficients are shown below as blue diamonds in Figure 5.
Figure 5: NACA 2412 airfoil Cl vs α with linear curve fit
The zero-lift angle of attack for this airfoil occurs at
α = −2.23◦
(d) Discussion of answer:
• The line produced is linear, matching what was discussed in theory during class.
• The camber of the airfoil produces positive lift even at zero angle of attack.
• The zero-lift angle of attack is negative because of the airfoil camber.
• Small changes of angle of attack result in large changes in lift, and these change
only occur in a small range of angles.
• An angle of attack of α = −5◦ results in a negative lift coefficient, which would
result in a lift vector that is pointed downwards.
14
(e) MATLAB code attached.
1 %{
2 Andrew Koehler
3 AE 202: Aerospace Flight Mechanics
4 Teaching Assistant | Spring 2020
5 Homework Assignment 3
6 Question 4
7 %}
8 %% setup
9 clear; close all; clc;
10
11 %% Given
12 % NACA 2412 airfoil
13 alpha = [0; 5; −5]; % angles of attack
14
15 %% Load data
16 naca_2412_0deg = csvread('naca2412i_1.csv',1,0);
17 naca_2412_5deg = csvread('naca2412i_2.csv',1,0);
18 naca_2412_neg_5deg = csvread('naca2412i_3.csv',1,0);
19
20 upper_0deg.xc = naca_2412_0deg(:,1);
21 upper_0deg.cp = naca_2412_0deg(:,2);
22 lower_0deg.xc = naca_2412_0deg(:,3);
23 lower_0deg.cp = naca_2412_0deg(:,4);
24
25 upper_5deg.xc = naca_2412_5deg(:,1);
26 upper_5deg.cp = naca_2412_5deg(:,2);
27 lower_5deg.xc = naca_2412_5deg(:,3);
28 lower_5deg.cp = naca_2412_5deg(:,4);
29
30 upper_neg_5deg.xc = naca_2412_neg_5deg(:,1);
31 upper_neg_5deg.cp = naca_2412_neg_5deg(:,2);
32 lower_neg_5deg.xc = naca_2412_neg_5deg(:,3);
33 lower_neg_5deg.cp = naca_2412_neg_5deg(:,4);
34
35 % get rid end of lower arrays which don't have airfoil data
36 lower_0deg.xc = lower_0deg.xc(1:78);
37 lower_0deg.cp = lower_0deg.cp(1:78);
38
39 lower_5deg.xc = lower_5deg.xc(1:78);
40 lower_5deg.cp = lower_5deg.cp(1:78);
41
42 lower_neg_5deg.xc = lower_neg_5deg.xc(1:78);
43 lower_neg_5deg.cp = lower_neg_5deg.cp(1:78);
44
45 %% Calculations
46 % Find Cl
47 Cl = zeros(3,1);
48 % without linear interpolation:
49 % Cl(1) = trapz(lower_0deg.xc,lower_0deg.cp) − trapz(upper_0deg.xc
,upper_0deg.cp);
50 % Cl(2) = trapz(lower_5deg.xc,lower_5deg.cp) − trapz(upper_5deg.xc
,upper_5deg.cp);
15
51 % Cl(3) = trapz(lower_neg_5deg.xc,lower_neg_5deg.cp) − trapz(
upper_neg_5deg.xc,upper_neg_5deg.cp);
52 %
53 n = 100;
54 x = linspace(1/n,1,n);
55
56 lower_0deg_interp = interp1(lower_0deg.xc,lower_0deg.cp,x);
57 upper_0deg_interp = interp1(upper_0deg.xc,upper_0deg.cp,x);
58
59 lower_5deg_interp = interp1(lower_5deg.xc,lower_5deg.cp,x);
60 upper_5deg_interp = interp1(upper_5deg.xc,upper_5deg.cp,x);
61
62 lower_neg_5deg_interp = interp1(lower_neg_5deg.xc,lower_neg_5deg.
cp,x);
63 upper_neg_5deg_interp = interp1(upper_neg_5deg.xc,upper_neg_5deg.
cp,x);
64
65 Cl(1) = trapz(x,lower_0deg_interp) − trapz(x,upper_0deg_interp);
66 Cl(2) = trapz(x,lower_5deg_interp) − trapz(x,upper_5deg_interp);
67 Cl(3) = trapz(x,lower_neg_5deg_interp) − trapz(x,
upper_neg_5deg_interp);
68
69 %% Plots
70 font = 18;
71 figure_size = [350 150 1200 750];
72 colors = [0 0.4470 0.7410; 0.8500 0.3250 0.0980; 0.9290 0.6940
0.1250; 0.4940 0.1840 0.5560; 0.4660 0.6740 0.1880; 0.3010
0.7450 0.9330; 0.6350 0.0780 0.1840];
73 set(groot,'defaulttextinterpreter','latex');
74 set(groot,'defaultAxesTickLabelInterpreter','latex');
75 set(groot,'defaultLegendInterpreter','latex');
76 mk_sz = 8;
77
78 %% figure 1
79 fig1 = figure(1);
80 plot(upper_0deg.xc,upper_0deg.cp,'Color',colors(1,:),'LineStyle','
−','Marker','o','MarkerSize',mk_sz,'Linewidth',2)
81 hold on
82 plot(lower_0deg.xc,lower_0deg.cp,'Color',colors(2,:),'LineStyle','
−','Marker','x','MarkerSize',mk_sz,'Linewidth',2)
83 hold on
84 grid on
85 set(gca,'FontSize', font)
86 set(gcf, 'Position', figure_size);
87 xlabel('$x/c, nd$')
88 ylabel('$C_{P}, nd$')
89 %title('$C_{P}$ vs $x/c$, $\alpha = 0^\circ$')
90 leg1 = legend('$C_{P}$ Upper','$C_{P}$ Lower','Location','
northeast');
91 leg1.FontSize = font;
92 set(gca, 'YDir','reverse')
93
94 %% figure 2
95 fig2 = figure(2);
16
96 plot(upper_5deg.xc,upper_5deg.cp,'Color',colors(1,:),'LineStyle','
−','Marker','o','MarkerSize',mk_sz,'Linewidth',2)
97 hold on
98 plot(lower_5deg.xc,lower_5deg.cp,'Color',colors(2,:),'LineStyle','
−','Marker','x','MarkerSize',mk_sz,'Linewidth',2)
99 hold on
100 grid on
101 set(gca,'FontSize', font)
102 set(gcf, 'Position', figure_size);
103 xlabel('$x/c, nd$')
104 ylabel('$C_{P}, nd$')
105 %title('$C_{P}$ vs $x/c$, $\alpha = 5^\circ$')
106 leg2 = legend('$C_{P}$ Upper','$C_{P}$ Lower','Location','
northeast');
107 leg2.FontSize = font;
108 set(gca, 'YDir','reverse')
109
110 %% figure 3
111 fig3 = figure(3);
112 plot(upper_neg_5deg.xc,upper_neg_5deg.cp,'Color',colors(1,:),'
LineStyle','−','Marker','o','MarkerSize',mk_sz,'Linewidth',2)
113 hold on
114 plot(lower_neg_5deg.xc,lower_neg_5deg.cp,'Color',colors(2,:),'
LineStyle','−','Marker','x','MarkerSize',mk_sz,'Linewidth',2)
115 hold on
116 grid on
117 set(gca,'FontSize', font)
118 set(gcf, 'Position', figure_size);
119 xlabel('$x/c, nd$')
120 ylabel('$C_{P}, nd$')
121 %title('$C_{P}$ vs $x/c$, $\alpha = −5^\circ$')
122 leg3 = legend('$C_{P}$ Upper','$C_{P}$ Lower','Location','
northeast');
123 leg3.FontSize = font;
124 set(gca, 'YDir','reverse')
125
126 %% figure 4
127 x1 = linspace(alpha(3), alpha(2))';
128 p1 = polyfit(alpha,Cl,1);
129 zero_lift_alpha = roots(p1);
130 y1 = polyval(p1,x1);
131
132 fig4 = figure(4);
133 scatter(alpha,Cl,150,colors(1,:),'d','filled')
134 hold on
135 plot(x1,y1,'Color',colors(2,:),'Linewidth',2)
136 grid on
137 set(gca,'FontSize', font)
138 set(gcf, 'Position', figure_size);
139 xlabel('$\alpha, ^\circ$')
140 ylabel('$C_{l}, nd$')
141 %title('$C_{l}$ vs $\alpha$')

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MAS8383 ECO 512 STAT 8310 CMN3102 MA577 CMPT 225 COMP0045 Methods Equations MATH 113 Math 3283W ICS-33 EN4302/ENT603 BIA B350F MA318 MTHM506 MECH 5315M ECONOMICS MAS8382 COM 2004 PX3142 XJEL 3662 STAT 231 MGT B399F PX4131 COM2108 30AM MSCI521 ECONM1015 ACFI304 GR5241 MATH10101 ECON 2070 COMS31900 EFIM20011 CSE 250B ACSE-5 STAT4207 SOFT2412 ELEC 331 AY2020 MSIN0209 FIT2094 CS 241L ST404 CS 411 3D CA3 INF4002 COMP3506 SIT718 CS 188 CS 170 MATH 138 CSE 251A CMT118 DATA5207 CMPT307 CMPT459 T1 PGEE11108 COSC 3P32 ME 163 MATH96007 7QQMM203 EE6227 BA 574 STA 4234 ECO206Y1Y ECO2008 MT2300 MATH 11158 CS 527 MACM 316 CSE474 2DI70 ISE 525 COMP3411 EMAT30007 7CCSMBDT MFIN704 MFIN704W21 MfIB ECE 438 CS 436 F36 CS5344 FE 621A FE621 TBA 612 ECM3151 ECMM422 MF1 MSIN0107 ESSMENT CS 5854 MFIT 5008 MIPS R4000 ENG5274 BU52018 ENG2079 BIOA02 STA2570 F0 INF553 STAT 440 STAT3008 COMP9417 COMP6451 6COM1042 BU52006 ST303 HW-1 CISC-235 CS 510 MATH3161 ECN 140 ECE391 Comp 3613 CVEN 9625 PRG455 UESTION 1 EARN 5 FE 825 IERG 5130 STAT 8178 EE161 ADS2 G54FUZ CS326 113B ECN 226 QBUS6860 ELEC3202 COMP9334 MGT11B MET CS 777 STA238 00099F Python SMM306 CMT206 EC116 SP MAST90084 CS480/680 BCPM0091 CMPT 300 ECON2209 FY20 STAT 432 CVEN9625 MATH1712 CMP9771M CMP3750M CIS 325 AI506 MAT344H1S MGT7180 PSTAT 160B CSYS5010 1D STAT 453/558 ME152B STA304H1F CSCI-128 ME428 CO3002 FINE 7640 COMP3234 ECSE 223 EE497 – 3D MATH5905 ECS648U ENG5009 MATH10242 EC665 A CPI221 ISOM5220 MACE40402 MGT B456F COMP9315 AFM 113 STAT 778 CS 564 MA231 CS 230 CS 124 COMP5511 CSC317 ITEC1010 ELEC207 INFT 3019 7CCMFM06 EE 161 5G ECLT5810 MATH 7349 CS526 O2 COMP90054 ELE00063M ACCT5034 ARC1015 CSCI 40 CSCI E40 FIN 524B CS 462 389COM CMT309 MFIN703 COMP4250 STA304 STA304H1S STA 1003 FIN 448 BUFN 732 CS520 CVEN3701 ECMT6006 PHAS0100 CSIT314 MDIA 5028 CSCI 2134 MAT 451 CMPT 459 STAT 457 INFO20003 COMP4141 IELE8066 F540 ECM9 CSE 6321 FIN 430 MATH3911 BFC5916 COMP1011 EEE 6209 STATS 240 CW3 COMP206 COMP1003 4331W ISE 530 CMPT 295 MATH 239 QF5203 INFR08010 IT114105 FIT2093 ECON3124 ECON1195 MATH 337 STAT 3006 COMP202 COMP3130 MATH3609 COMP 250 CS6140 IEEE ELE00041I FIN40090 CIV 5899 MATH3066 IGNMENT MATH 2009 STAT3405 HIST 101 MATH 3281 CEGE0042 ICT532 LAB 7 LAB 5 MCO455 CSIT214 Q4 IFB104 A1 ECO220Y5Y DA5020 CMPSC473 FIN9007 DSCI553 MSCI534 COMP3308 CSE-112 CMPT 310 EE104 F 2C P20 ALY2010 BUSM060 STA258 ECO00030M R BIOSTAT 274 COS6012-B FIT3155 CMPUT 340 EC421 INFS4205 ECOM135 NATS 1740 COMP809 MBE 6005 STA465 CSSE2310 CIVL2700 DATA4700 CMPSC 473 ISE 529 CS5044 MAST30020 COMP2207 ENGG1001 COMP0046 FIT3165 STAT GR5241 MECH0007 vATHK1001 ANALYTIC ATHK1001 MATH352001 HW3 S1 BMAN 24662 COMP 636 Stat-461 STAT 302 STAT 2132 STAR534 ENG4137 CMP3752M CS3103 BISM3222 STA 135 ME155A IIMT3601 ECON3371 ECON 3371 HW4 FINA 5840 DT211C CTEC3902 MAT005 MATH2148 S261F Econ 332 MATH3734 MIS41270 7CCMFM13 SYST 664 CIS434 CENG10014 MT4111 MT4527 ECMM424 CMT219 Math 136 MATH 3120 COMP10002 MAST20009 MATH0058 BEEM117 ISYE 6420 MGMTMFE 405 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