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AAE 451 AERODYNAMICS PDR 2 TEAM 4 Jared Hutter, Andrew Faust, Matt Bagg, Tony Bradford, Arun Padmanabhan, Gerald Lo, Kelvin Seah November 18, 2003

AERODYNAMICS PDR 2

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AERODYNAMICS PDR 2. TEAM 4 Jared Hutter, Andrew Faust, Matt Bagg, Tony Bradford, Arun Padmanabhan, Gerald Lo, Kelvin Seah November 18, 2003. OVERVIEW. Concept Review Aircraft CL and CM Updated Wing Size Aircraft Plots Follow-Up Actions. CONCEPT REVIEW. Empennage - PowerPoint PPT Presentation

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Page 1: AERODYNAMICS PDR 2

AAE 451

AERODYNAMICSPDR 2

TEAM 4Jared Hutter, Andrew Faust, Matt Bagg, Tony Bradford,Arun Padmanabhan, Gerald Lo, Kelvin Seah

November 18, 2003

Page 2: AERODYNAMICS PDR 2

TEAM4OVERVIEW

Concept Review

Aircraft CL and CM

Updated Wing Size

Aircraft Plots

Follow-Up Actions

Page 3: AERODYNAMICS PDR 2

TEAM4CONCEPT REVIEW

High WingS = 47.8 ft2

b = 15.5 ft, c = 3.1 ftAR = 5

Twin Booms3 ft apart;

7.3 ft from Wing MAC to HT MAC

Twin Engine1.8 HP each

Avionics Pod20 lb; can be positioned front or aft depending on

requirements

EmpennageHorizontal and Vertical

Tails sized using modified Class 1 Approach(per D & C QDR 1)

Page 4: AERODYNAMICS PDR 2

TEAM4AIRCRAFT LIFT COEFFICIENT

Lift Coefficient

CL = CL* + CLe*elevator + CL0

Matlab script based on Roskam Vol VI Ch8:

CL = 5.41(rad-1)* + 0.4675(rad-1)*elevator + 0.3086

Predator Codes from AAE 565

CL = 5.473(rad-1)* + 0.454(rad-1)*elevator + 0.3113

Page 5: AERODYNAMICS PDR 2

TEAM4AIRCRAFT PITCHING MOMENT

Moment Coefficient

CM = CM* + CMe*elevator + CM0

Matlab Script based on Roskam

CM = -2.0496 (rad-1)* + (-0.1771)(rad-1)* elevator + 0.0425

Predator Codes

CM = -2.2682 (rad-1)* + (-1.058)(rad-1)* elevator - 0.2785

Page 6: AERODYNAMICS PDR 2

TEAM4AIRCRAFT CL AND CM

AAE 565 Predator code similar to Roskam

Roskam uses graphs in his book

Predator has the graphs hard coded into the program

Predator will be more accurate

Update Constraint Diagram

Need Maximum CL for Constraint Diagram

Roskam Code solves for Maximum CL

.06 difference between Roskam Code and Predator for CL

Page 7: AERODYNAMICS PDR 2

TEAM4

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.05

0.1

0.15

0.2

0.25

0.3

0.35Distribution of section cl along span of the wing @ AOA=0

Percent Span (y/(b/2))

Sec

tion

cl

MAXIMUM LIFT COEFFICIENT

Need section cl along wing span

Increase Angle of Attack and find new section cls

Repeat until the wing begins to Stall

That is the stall angle Integrate section cl’s to find

Maximum CL

Page 8: AERODYNAMICS PDR 2

TEAM4

AIRCRAFT CL AND CM

Three Major Codes: Predator, CL max, and Constraint Diagram Predator:

Input: Aircraft Geometry Output: CL and CM equations

Maximum Lift Coefficient Input: Main Wing and Horizontal Tail Geometry Output: CL Max and Alpha at CL Max

Constraint Diagram Input: Flight Conditions, CL at 0 Alpha, CL max, Engine Info Output: Wing Area and Required Power

Page 9: AERODYNAMICS PDR 2

TEAM4

AIRCRAFT CL AND CM

Iterative loop can be used

Used old constraint diagram values for initial guess

Used Wing Area as the Control variable

Constraint Code

Predator Code

Max CL Code

Page 10: AERODYNAMICS PDR 2

TEAM4

AIRCRAFT CL AND CM

Lift Coefficient

CL = CL* + CLe*elevator + CL0

CL = 5.931(rad-1)* + 0.59(rad-1)*elevator + 0.2809

Moment Coefficient

CM = CM* + CMe*elevator + CM0

CM = -3.6947(rad-1)* + (-1.058)(rad-1)* elevator -.3956

Reduce CM0 for clean flight

Page 11: AERODYNAMICS PDR 2

TEAM4

AIRCRAFT CL AND CM

CM0 main contribution is from the Incidence angle of Horizontal Tail (-2.51 degrees)

Using the Iterative Loop, ran over a range of Horizontal Tail Incident angles

Found Incident Angle that reduced CM0 the most

Page 12: AERODYNAMICS PDR 2

TEAM4

AIRCRAFT CL AND CM Lift Coefficient

CL = CL* + CLe*elevator + CL0

CL = 6.0339(rad-1)* + 0.6201(rad-1)*elevator + 0.4237

Moment Coefficient

CM = CM* + CMe*elevator + CM0

CM = -4.0421(rad-1)* + (-1.058)(rad-1)* elevator + 0.00

Incident Angle=.23 degrees

Does not seem right, may be caused by Downwash from the Main Wing

Page 13: AERODYNAMICS PDR 2

TEAM4

AIRCRAFT PARAMETERS

Wing Area= 34.5 ft^2 Wing Span= 13.1 ft Max CL= 1.8034 @ 12.88 Degree Angle of Attack CD= .0339 @ 0 Angle of Attack

Page 14: AERODYNAMICS PDR 2

TEAM4

-10 -5 0 5 10-1

-0.5

0

0.5

1

1.5

2

Angle of Attack (degrees)

CL

-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

1.5

2

CM

CL

Dele 10d

Dele 0d

Dele -10d

TRIM DIAGRAM AT CRUISE

CL=.4327

Page 15: AERODYNAMICS PDR 2

TEAM4

0.02 0.04 0.06 0.08 0.1 0.12 0.14-0.5

0

0.5

1

1.5Drag Polar at Cruise

CD

CL

Drag Polar

Based on Roskam Vol VI Ch 4

Page 16: AERODYNAMICS PDR 2

TEAM4FOLLOW-UP ACTIONS

Verify CD calculations

Triple Check CM0 and Incident Angle of the

Horizontal Tail

React to changes from D+C, Propulsion,

and Structures

Page 17: AERODYNAMICS PDR 2

AAE 451

Questions?

Page 18: AERODYNAMICS PDR 2

TEAM4

AppendixLift Curve Slope CL = f(CLW, CLHT, HT, w) HT = Ratio of dynamic pressure. Mostly caused by

propeller wash and velocity Downwash, w = Caused by main wing’s vortex flow on tail. Changes effective angle of attack for the tail.

elevatorPositive

Negative

Page 19: AERODYNAMICS PDR 2

TEAM4AIRCRAFT PARAMETERS Lift Curve Slope for Elevator Deflection

CLe = f(elevator size, horizontal tail planform)

Zero Angle of Attack Lift Coefficient

CL0 = f(CL0W, CL0HT, HT, incident angles)

HT = Ratio of dynamic pressure. Mostly caused by

propeller wash and velocity

Incident angles are for both main wing and

horizontal tail

Page 20: AERODYNAMICS PDR 2

TEAM4AIRCRAFT PARAMETERS Moment Coefficient

CM = CM* + CMe*elevator + CM0

CM = -0. 0225(deg-1)* + (-0.0027)(deg-1)* elevator + 0.0280

Moment Curve Slope

CM = f(dCM/dCL, CL)

dCM/dCL = f(CG, Aerodynamic Center of Aircraft)

Page 21: AERODYNAMICS PDR 2

TEAM4AIRCRAFT PARAMETERS Zero Angle of Attack Moment Coefficient

CM0 = f(CM0_W, CM0_HT [both about the CG])

LIFT

WEIGHT

Aerodynamic Center