Final Design Presentation -- Aerodynamics

8/8/2019 Final Design Presentation -- Aerodynamics

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Aerodynamics

Final Design Report


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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


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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


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AerodynamicsObjectives

Minimize Drag

Minimize Craft Size

Maximize Lift

Structural Integrity

A Stable and Self Righting Craft


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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.


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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


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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


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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


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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


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AerodynamicsExcel Screenshot


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AerodynamicsMatlab Screenshots


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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


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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


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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


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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


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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


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AerodynamicsSelig Airfoil gm15


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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


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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


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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.jpg


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AerodynamicsTwo Sample Wings


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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


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Aerodynamics Performance

Stall Speed

Velocity of Max Lift to Drag Ratio

smSC

WV

L

stall /832

max

smeARCS

WV

DP

DL /1202

max/


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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


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Stability & ControlLongitudinal

Moment Coefficient Cm

ScV

MC cgM 2

21


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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/


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Moment Contribution from the wing

(z ~ 0) (Dw


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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


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Moment Contribution from the Tail

lt ~ 1 m

it ~ 3o

0

;momentspreviousbalanceto194.0

o

o

tth iV

e

ea

ww

tth

Sc

SlV


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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


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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


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Stability & ControlTail Shape Selection

Traditional

V-shape

Inverted V

Inverted Y Fork Shape


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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


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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.


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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


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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


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Stability & ControlLateral & Directional

Dihedral angle

Generates Roll Stability

3.5o

Sweepback Generates Yaw Stability

5o


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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


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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

Documents

Final Design Presentation -- Aerodynamics