View
218
Download
0
Category
Preview:
Citation preview
8/8/2019 Final Design Presentation -- Aerodynamics
1/40
Aerodynamics
Final Design Report
8/8/2019 Final Design Presentation -- Aerodynamics
2/40
Weight
Airfoil Shape
Low Reynolds Number Problems
Boundary layer is much less capable of handling adversepressure gradient without separation
Laminar flow: laminar separation bubbles can lead to
excessive drag and low maximum lift
Stability Tail Design
Fuselage Design
AerodynamicsConsiderations
8/8/2019 Final Design Presentation -- Aerodynamics
3/40
Laminar Flow (low Reynolds number)
High speed Greater Drag
Small wing area Low lift
Low lift coefficient due to low Martian density
Low Aspect Ratio high induced drag
Lower Speed of sound on Mars from lowTemperature
AerodynamicsProblems & Concerns
8/8/2019 Final Design Presentation -- Aerodynamics
4/40
AerodynamicsObjectives
Minimize Drag
Minimize Craft Size
Maximize Lift
Structural Integrity
A Stable and Self Righting Craft
8/8/2019 Final Design Presentation -- Aerodynamics
5/40
AerodynamicsWing Design Process
Choose approximate Cl value based off excelsheet.
Find an airfoil with appropriate Cl and Cd
values using Matlab
Use Excel to build an, optimized planformwith inputted constraints
Use coordinates generated in Matlab andexcel to build wing in ProE.
8/8/2019 Final Design Presentation -- Aerodynamics
6/40
AerodynamicsAirfoil Selection
Need a low Re airfoil for Martian Atmosphere
Airfoil data obtained from Michael Selig ofUniversity of Illinois
Data Analyzed using Matlab and Excel.
Matlab for plotting and number crunching
Excel for generating planform
8/8/2019 Final Design Presentation -- Aerodynamics
7/40
Re must match data from Selig files Freestream mach number must be kept below
0.6 for propeller tip velocity to remain subsonic
AR must be chosen to minimize drag
Structural Stability must be kept in mind whiledesigning a planform
AerodynamicsPlanform Constraints
8/8/2019 Final Design Presentation -- Aerodynamics
8/40
AerodynamicsAssumptions
Density, temperature and viscosity gradientsare as defined by empirical equations
Selig airfoil data is correct and reliable
Planform will be smooth
Controller can be designed to stabilize craft
Assume relatively tranquil flying conditions
Valid due to predictable Martian seasons
8/8/2019 Final Design Presentation -- Aerodynamics
9/40
Atmospheric Conditions
Average Temperature = 223 K = -50C
Pressure = 700 Pa
Density = 1.57 e-2 kg/m3
g = 1.29
Speed of sound = 234.72 m/s
8/8/2019 Final Design Presentation -- Aerodynamics
10/40
AerodynamicsExcel Screenshot
8/8/2019 Final Design Presentation -- Aerodynamics
11/40
AerodynamicsMatlab Screenshots
8/8/2019 Final Design Presentation -- Aerodynamics
12/40
AerodynamicsCL Selection
First we needed to specify a desired CL
This is done based on Reynolds number
As the velocity increases Reynolds numberincrease and the lift obtained goes upRequired CL decreases
The opposite is the same:
As velocity decreases the CL needed increasesand the Reynolds number decreases
8/8/2019 Final Design Presentation -- Aerodynamics
13/40
AerodynamicsRe vs. Optimal CLOptimal Cl v. Reynolds Number
35000
40000
45000
50000
55000
60000
65000
70000
0.3 0.35 0.4 0.45 0.5 0.55 0.6
Cl
Re
8/8/2019 Final Design Presentation -- Aerodynamics
14/40
AerodynamicsCL Choice
After careful analysis we have chosen to flyat a lower Reynolds number to minimize drag
To do this we need greater CL So we selected a target CL of 0.55
8/8/2019 Final Design Presentation -- Aerodynamics
15/40
Data For 6 optimized wings
wing drag weight b sweep Cr Ct Vel
gm15 2.4442 0.82 2.078 4.9 0.422 0.084 129.1
s6063 2.9398 1.86 3.100 7.1 0.966 0.193 83.4
s7012 2.5381 1.41 2.922 4.9 0.627 0.125 129.1
rg14 2.8761 2.22 3.645 7.1 1.131 0.226 71.69
rg14 2.8331 1.33 2.848 5.0 0.628 0.126 129.1
gm15sm 2.6638 2.55 4.000 6.2 1.087 0.217 74.97
gm15sm 2.4501 1.59 3.162 4.6 0.631 0.126 129.1
sd7003 2.5914 2.62 4.000 6.6 1.164 0.233 70.4
sd7003 2.5313 2.34 4.000 4.8 0.840 0.168 97.6
8/8/2019 Final Design Presentation -- Aerodynamics
16/40
Wing Geometry
Wing Twist: prevent tip stall and to revise thelift distribution to approximate an ellipse
Taper Ratio: Ratio between the tip and the
root chord; affects the lift distribution of thewing
Aspect Ratio: b2/Sor b/cbar High AR --> Lower Induced drag for same area
Sweep
Dihedral AngleDiscussed later with stability
8/8/2019 Final Design Presentation -- Aerodynamics
17/40
AerodynamicsSelig Airfoil gm15
8/8/2019 Final Design Presentation -- Aerodynamics
18/40
Aerodynamics -- Wing Position
High
Low Fuselage used in military transport
Increases frontal area and thus drag
Low
Landing Gear Storage
Dihedral angle not set by aerodynamics will
increase the size of tail Mid
Better Maneuverability
8/8/2019 Final Design Presentation -- Aerodynamics
19/40
AerodynamicsWing Data
b = 2.07 m
Croot = 0.432 m
Ctip = 0.1512 m
Taper = 0.35 AR = 7.2
M.A.C = 0.307 m
Design Cl = 0.58
M = 0.56 t/c = .087
Re = 40600
Sweep = 5o
Dihedral Angle = 3.5o
Wing Twist = 3o
8/8/2019 Final Design Presentation -- Aerodynamics
20/40
AerodynamicsWing Tip Selection
Pros:
Prevents vortex around wing tip which decreaseslift at the tip
Serves same purpose of winglet withoutincreasing the wetted area
http://www.modellbau-pollack.de/shop/f3j/images/eraserf3j-winglet.jpg8/8/2019 Final Design Presentation -- Aerodynamics
21/40
AerodynamicsTwo Sample Wings
8/8/2019 Final Design Presentation -- Aerodynamics
22/40
AerodynamicsFuselage Design
We based our Fuselage Design on that of a Sailplane
WHY?
No boundary Layer separation points on the fuselage
Converged rear prevents vortex wake from forming Small wetted area
8/8/2019 Final Design Presentation -- Aerodynamics
23/40
Aerodynamics Performance
Stall Speed
Velocity of Max Lift to Drag Ratio
smSC
WV
L
stall /832
max
smeARCS
WV
DP
DL /1202
max/
8/8/2019 Final Design Presentation -- Aerodynamics
24/40
Stability & Control - Longitudinal
Make an assumption of where the center of gravityis going to be
Make an educated guess of the placement of the tail
and the wing Determine the moment about the C.G due to wing,
fuselage, tail, and payload
If Cm (L=0) is positive and dcm/da is negative then
UAV is longitudally and statically stable Iterate the process if necessary
8/8/2019 Final Design Presentation -- Aerodynamics
25/40
Stability & ControlLongitudinal
Moment Coefficient Cm
ScV
MC cgM 2
21
8/8/2019 Final Design Presentation -- Aerodynamics
26/40
Stability & ControlLongitudinal
1) (Cm,cg)L=0 = Cm,acwb + VHat(it+eo) + Cm(payload) Cm,acwb is the moment contribution from the wing and the fuselage
VHat(it+eo) is the moment contribution from the tail
Cm(payload) is the moment contribution from the payload
We want (Cm,cg)L=0 to be positive
2) We want this term to be as more negative as possible. The
smaller the value, the more stable the UAV.
a
eaaa
a d
dVhh
d
dCmthacwb 1/
8/8/2019 Final Design Presentation -- Aerodynamics
27/40
Moment Contribution from the wing
(z ~ 0) (Dw
8/8/2019 Final Design Presentation -- Aerodynamics
28/40
Stability & Control - Longitudinal
Moment Contribution for the payload
Total Moment For Payload = -3.733 Nm
Moment Coefficient = -0.153
0m 2mcg
0.85
0.75
0.650.25 1.60
1.75
0.61kg
Camera
4.2kg
Battery
1.17kg
Lidar
Controls
1.8kg
Motor 0.43kg
drive
shaft
0.6kg
Propeller
8/8/2019 Final Design Presentation -- Aerodynamics
29/40
Stability & Control - Longitudinal
Moment Contribution from the Tail
lt ~ 1 m
it ~ 3o
0
;momentspreviousbalanceto194.0
o
o
tth iV
e
ea
ww
tth
Sc
SlV
8/8/2019 Final Design Presentation -- Aerodynamics
30/40
Stability & Control - Longitudinal
Is the UAV stable?
(Cm,cg)L=0 = Cm,acwb + VHat(it+eo) + Cm(payload)=0.05
Both of the criteria are satisfied, therefore the UAV
is longitudinally stable
03.01/
a
eaaaa d
dVhhddCm thacwb
8/8/2019 Final Design Presentation -- Aerodynamics
31/40
Stability & ControlLongitudinal
Neutral Point
It is a fixed point on the UAV behind the center ofgravity where is equal to zero
At this position the UAV is considered to betrimmed
Solving for hn (the location of the neutral point)yields 1.06m back from nose
Static Margin: hn-hcg = 0.21m
Larger the value the more stable the craft is
ad
dCm
8/8/2019 Final Design Presentation -- Aerodynamics
32/40
Stability & ControlTail Shape Selection
Traditional
V-shape
Inverted V
Inverted Y Fork Shape
8/8/2019 Final Design Presentation -- Aerodynamics
33/40
Stability & Control - Tail
Determining the tail configuration
Pick Lh (distance between tail and cg)
Lh determines the tail surface area
Tradeoff between stability and drag
Airfoil Selection
Decide on a AR
Determine the chords Decide on a tail setting angle
Iterate if necessary
8/8/2019 Final Design Presentation -- Aerodynamics
34/40
Stability & Control - Tail
Tail airfoil candidates
Airfoil at Cl/Cd @ it = 3o
Rg14 .1051 27
S7012 0.0982 23S7003 .0993 38
S6063 .0933 20
We chose S7003 because it has a positive lift coefficient at zero
angle of attack. In addition, its lift to drag ratio at it=3 is greaterthan the other three candidates.
8/8/2019 Final Design Presentation -- Aerodynamics
35/40
Stability & Control - Tail
The desired operating Reynolds number is61100. However, it is ok to operate at alower Re, because the tail angle of attack is
relatively small; therefore there will be littleboundary layer separation
Pick an AR ~ 4 Too large the chord will be excessively small,
which leads to super low Reynolds number Too small there will be a substantial increase in
induced drag from the tail
8/8/2019 Final Design Presentation -- Aerodynamics
36/40
Stability & Control - Tail
St: 0.15 m2
Aspect ratio: 4
Taper ratio: 0.35
Root Chord: 0.287 m
Tip Chord: 0.10 m
M.A.C: 0.209 m
Reynolds number: 29,375
8/8/2019 Final Design Presentation -- Aerodynamics
37/40
8/8/2019 Final Design Presentation -- Aerodynamics
38/40
Stability & ControlLateral & Directional
Dihedral angle
Generates Roll Stability
3.5o
Sweepback Generates Yaw Stability
5o
8/8/2019 Final Design Presentation -- Aerodynamics
39/40
Stability & ControlControl Surfaces
Ailerons
Approximately 50-90% of the wing span
Approximately 20% of wing chord
Rudders & Elevators Since we have an inverted V-Tail the Rudders and
Elevators in the tail are combined
Final Design needed
8/8/2019 Final Design Presentation -- Aerodynamics
40/40
Stability & Control
Stick fixed vs. Stick free stability control
Since the craft is a UAV it is stick fixed
This means that all elevators and rudder control
as well as aileron control are going to becontrolled by computers with sensors
Recommended