Aerodynamics PDR AAE451 – Team 3 October 21, 2003

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Aerodynamics PDR AAE451 – Team 3 October 21, 2003 Brian Chesko Brian Hronchek Ted Light Doug Mousseau Brent Robbins Emil Tchilian

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Aerodynamics PDR AAE451 – Team 3 October 21, 2003. Brian Chesko Brian Hronchek Ted Light Doug Mousseau Brent Robbins Emil Tchilian. Aerodynamics PDR. Design Process Span-wise distribution for c l found using Lifting-Line Theory Airfoil Selection Drag Integration. Lifting-Line Theory. - PowerPoint PPT Presentation

Transcript of Aerodynamics PDR AAE451 – Team 3 October 21, 2003

Page 1: Aerodynamics PDR AAE451 – Team 3 October 21, 2003

Aerodynamics PDR

AAE451 – Team 3

October 21, 2003

Brian CheskoBrian HronchekTed LightDoug MousseauBrent RobbinsEmil Tchilian

Page 2: Aerodynamics PDR AAE451 – Team 3 October 21, 2003

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AAE 451Team 3Team 3 Aerodynamics PDR

Design Process

• Span-wise distribution for cl found using Lifting-Line Theory

• Airfoil Selection

• Drag Integration

Page 3: Aerodynamics PDR AAE451 – Team 3 October 21, 2003

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AAE 451Team 3Team 3 Lifting-Line Theory

Inputs:a = dcl/d5.66 (airfoil specific)

Constant Chord = 2.8

AR = 5

= 6.85 deg (to match CL)

(actually L=0)

• Wing modeled as distribution of horseshoe vortices

• Fourier Series for circulation along span

nAbV

N

nn sin2)(

1

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AAE 451Team 3Team 3 Prandtl’s Lifting-Line

• Solve Prandtl’s wing equation

• Substitute

iLl

cVaa

c

000

2

nAbVN

nn sin2)(

1

sin

sin)(

1

nnA

N

nni

0110 sin

sinsin

4

L

N

nn

N

nn

nnAnA

ca

b

•System of N equations with N unknowns (Solve N N matix)

•Take N different spanwise locations on the wing where the equation is to be satisfied: 1, 2, .. N; (but not at the tips, so: 0 < 1 < )

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AAE 451Team 3Team 3 Lifting-Line For Rectangular Wing

• Consider: rectangular wing: c = constant; span = b; b/c = A = 5 without twist: = constant; L=0 = 0• Evaluate the wing equation at the N control points at i :

• The wing is symmetrical A2, A4,… are zero– take only A1, A3,… as unknowns– take only control points on half of the wing: 0 < i /2

• Example for N=30:– take A1, A3, A5 as unknowns– take control points (equidistant in ): = /(2N) stepping– take lift-slope of the airfoil a0 = 5.66, and wing aspect ratio A = 5

i

N

nn

i

nAn

a

Asin

sin

4

1 0

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AAE 451Team 3Team 3 Lifting-Line Calculation

• Sample Output

N = 3

• CL calculation from lifting line theory

CL = πAR*A1*α CL = W/S*q = .4873 from constraint

solve for = 6.8 deg in order to match CL

• CDi calculation

• cl calculation

1

1

1

975

464.80464.4

7103

3

2

1

A

A

A

0040.0

0277.0

2316.0

3

2

1

A

A

A

0where)1(2

2

1

2

N

n

nLD A

An

A

CC

i

nAbVN

nn sin2)(

1

iL

l

cVaa

c

000

2

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AAE 451Team 3Team 3 Lifting-Line Theory

Outputs:

CDi = 0.0131

Cdi distribution

CL = 0.4873

Cl distribution

Section Lift Coefficient Varies from ~ 0 – 0.6

0 1 2 3 4 5 6 70

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Span/2 (ft)

Cl,

Cdi

, al

phai

ClCd*10alpha

i*10

Page 8: Aerodynamics PDR AAE451 – Team 3 October 21, 2003

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AAE 451Team 3Team 3 Airfoil Selection

Airfoils Selection Criteria:

• Low drag over range of specified cl values

• Easy construction– Round Leading Edge– Relatively Flat Bottom

• Easy to construct on tabletop

– Constructible Trailing Edge

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AAE 451Team 3Team 3

-1 -0.5 0 0.5 1 1.50.005

0.006

0.007

0.008

0.009

0.01

0.011

0.012

0.013

0.014

0.015

Section Lift Coefficient, cl

Sec

tion

Dra

g C

oeff

icie

nt,

c d

Drag Polar for Candidate Airfoils

NACA 441244154425241823018ClarkY

Airfoil Selection

Region of Interest

Clark Y Airfoil is Best

Clark Y

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AAE 451Team 3Team 3 Clark Y Airfoil

-10 -8 -6 -4 -2 0 2 4 6 8 10-1

-0.5

0

0.5

1

1.5

2

Angle of Attack

Sec

tion

Lift

Coe

ffic

ient

, c l

-0.2

-0.1

0

0.1

0.2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x/c

y/c

•Geometry

•Drag Polar

•cl vs

-1 -0.5 0 0.5 1 1.5 20.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

0.05

Section Lift Coefficient, cl

Sec

tion

Dra

g C

oeff

icie

nt,

c d

dcl/d = 5.66

cl vs. cd vs. cl

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AAE 451Team 3Team 3 Total Lift and Drag Coefficient Estimation

• Lift:• CL – Found at cruise, can be obtained at any

• cl - Found using lifting line theory

• Drag:• CD = CDi + CDp

• CDi found using lifting line theory, can be obtained at any

• From Drag Polar of airfoil (cl vs. cd), cdp can be obtained and integrated to obtain CDp for the entire wing

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AAE 451Team 3Team 3 Parasitic Drag Calculation

• Used Polynomial Function to Fit Airfoil’s Drag Polar

-1 -0.5 0 0.5 1 1.5 20.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

0.05

0.055

Cl

Cd

AIRFOIL DRAG POLAR

ActualPolynomial Fit

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AAE 451Team 3Team 3 Parasitic Drag Calculation

• Plugged wing cl distribution into polynomial function to get corresponding parasitic cd distribution along span

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AAE 451Team 3Team 3 Parasitic Drag Calculation

• Integrated Parasitic Drag Distribution Along Span to get 3-D Wing Parasitic Drag

• CDp = .0059

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AAE 451Team 3Team 3 Total Wing Drag Coefficient

• CD = CDi + CDp

• CD = .0131 + .0059 = .0190

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AAE 451Team 3Team 3 Wing Characteristics

Wing Sweep = 0º

Taper Ratio = 1

Dihedral Angle = 5º

AR = 5

S = 40 ft2

Tail Airfoil = NACA 0012

(empirically based from Roskam Part II, p. 154)

(subject of future trade study)

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AAE 451Team 3Team 3 Wing Characteristics

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AAE 451Team 3Team 3 Coming Attractions…

• CLmax

• Control Surface Sizing

• Tail Sizing

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AAE 451Team 3Team 3 Questions?