Airfoils Wings

NACA Airfoils

6-Feb-08

AE 315 Lesson 10: Airfoil nomenclature and properties

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Definitions: Airfoil Geometry z Mean camber line

x Chord line Chord x=0 Leading edge

x=c Trailing edge

NACA Nomenclature

NACA 2421 1st Digit: Maximum camber is 2% of 2D airfoil chord length, c (or 3D wing mean chord length, c). 2nd Digit: Location of maximum camber is at 4/10ths (or 40%) of the chord line, from the LE. 3rd & 4th Digits: Maximum thickness is 21% of c (or c). 0.21 m 0.40 m 0.02 m

c=1m

Lift-Curve Slope Terminology c c l l

3

2

1

Sample NACA Data

α

4

[1] αl=0 – Angle of attack (α) where the lift coefficient (c l) = 0, ∴ no lift is produced; αl=0 = 0 for a symmetric airfoil; αl=0 < 0 for a positively cambered airfoil. [2] clα – Lift-curve slope (dc l/dα); ‘rise’ of cl over ‘run’ of α for a linear portion of the plot; ≈ 0.11/deg for a 2D (thin) airfoil (sometimes called a0) [3] clmax – Maximum cl the airfoil can produce prior to stall [4] αstall – Stall angle of attack; α at clmax; maximum α prior to stall

DATA SHOWN ON NACA CHARTS (2421)

Airfoil Shape

Data point symbols for various Reynolds numbers (R)

Location of aerodynamic center (a.c.)

Fill in the blanks: think – pair - share

NACA 6716 1st Digit: Maximum camber is _% of 2D airfoil chord length, c 2nd Digit: Location of maximum camber is at __ /10ths of the chord line, from the LE. NACA 6716 3rd & 4th Digits: Maximum thickness is _% of c). 0.14 0.09 c / 0.04 y

-0.01 -0.06 0

0.1

0.2

0.3

0.4

0.5

x/c

0.6

0.7

0.8

0.9

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Definitions: Symmetrical Airfoil If an airfoil is symmetrical, then the chord line is the line of symmetry. The part of the airfoil above the chord line will be the mirror image of the part below it.

Fill in the blank: If max thickness is 15% of chord – What is NACA 4-digit description?

Wing Geometry

) b ( n a S p Planform Area (S) S=b*c

Chord (c)

Angle of Attack, Airfoil Forces and Moments Lift

Aerodynamic Force

Moment +

V ∞

Drag

α

Relative wind

Angle of attack (

: angle between relative wind and chord line

Note: 1) lift is perpendicular to freestream velocity 2) drag is parallel to freestream velocity 3) moment is positive nose-up

Pressure & Shear Distribution Change with a Change in Angle of Attack Aerodynamic Force Vector – Sum of Pressure and Shear forces – Lift (perpendicular to flow) results mostly from pressure forces – Drag (parallel to flow) contains shear Lift and pressure forces

Total Aerodynamic Force (Sum of Pressure and Shear)

Pressure

Vfree stream Drag

Shear Figure 3.10 , Page 75

Center of Pressure Lift

Aerodynamic Force

Moment = 0 +

V ∞

Drag

α

Center of Pressure: the point on the airfoil where the total moment due to aerodynamic forces is zero (for a given α and V ) ∞

Increasing angle of attack causes the center of pressure to move forward while decreasing angle of attack moves th e center of pressure backwards.

Aerodynamic Center y

x

V ∞

M ac +

α

Aerodynamic Center : The point on the airfoil where the moment is independent of angle of attack. Fixed for •subsonic flight ≈ c/4 •supersonic flight ≈ c/2. The moment has a •negative value for positively cambered airfoils •zero for symmetric airfoils.

Force and Moment Coefficients l

Lift:

cl ≡

Drag:

cd ≡

Moment:

cm ≡

q ∞ S d q ∞ S

Note: nondimensional coefficients!

m q ∞ Sc

Coefficients for NACA airfoils are found from charts in Appendix B. Why are these coefficients a function of angle of attack and Reynolds number?


Airfoil Shape

Data point symbols for various Reynolds numbers (R)

Location of aerodynamic center (a.c.)


Lift Curve : cl plotted against α


Drag Polar: cd plotted against cl


Pitching moment coefficient at the quarter-chord point (cmc/4) plotted against α


Pitching moment coefficient at the aerodynamic center (cmac) plotted against cl

Daily quiz

Given:

NACA 2421 Airfoil Reynolds Number = 5.9x106 Angle of Attack = 12°

Find:

cl = clα = (∆ cl / ∆α) = cd = cm c/4 = cm a.c.= clmax = αstall = αl=0 =

Example Problem (NACA 2421)

cl ≈ 1.3

cl α =

0.6/6° = 0.1/°

Cm c/4 ≈ -0.025

Reynolds Number


cd ≈ 0.018 cl ≈ 1.3

Cm a.c. ≈ -0.04

Reynolds Number


clmax ≈ 1.4

αl=0 ≈ -2°

αstall ≈ 15°

Reynolds Number

Example Problem We just found: cl = 1.3 ; cd = 0.018 ; cmac = -0.04 To calculate the lift, drag and pitching moment on the airfoil we need to know the dynamic pressure, the chord, and the planform area. Given that we are at sea level on a standard day with V ∞ = 100 ft/sec, q = ½ ρ∞ V∞2 = ½ (0.002377 slug/ft3) (100 ft/sec)2 = 11.885 lb/ft2 If c = 4 ft and S = 200 ft 2,

l = cl q S = (1.3) (11.885 lb/ft2) (200 ft2) = 3090 lb d = cd q S = (0.018) (11.885 lb/ft2) (200 ft2) = 42.8 lb mac = cmac q S c = (-0.04) (11.885 lb/ft2) (200 ft2) (4 ft) = - 380 ft-lb

Changes to Lift Curves 1. Camber

cl

Positive camber

cl

Zero Camber (symmetric) Negative camber

α

c

2. 2D vs 3D wings

cl C L cl CL

Airfoil (2-D)

l 57 . 3 c

=

1+

Wing (3-D) α

6-Feb-08

C L

C L

= C L


l eAR

(

−

L = 0

) 24

Changes to Lift Curves 3. Flaps

cl

Without flaps With flaps

α

4. Boundary Layer Control (BLC) or increasing Reynolds Number

cl

Without BLC With BLC

α 6-Feb-08


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3-D Effects on Lift

c l and C L Airfoil

cl CL Wing

Notice the slope is decreased for the wing and the zero lift angle of attack is unchanged. 6-Feb-08


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Airfoils Wings

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