Aero Engineering 315

Lesson 13

Airfoils Part II

Where does the moment come from?

PU = Upper surface pressure distributionAupper

dAP force surfaceUpper U

PL = Lower surface pressure distribution lower

L

A

dAP force surfaceLower

Fnet

M

Note: M is negative for this exampleIn general:

M is < 0 for positive camberM is = 0 for symmetric airfoilsM is > 0 for negative camber

Note: Shear stress also contributes to moment in the same manner…

Center of Pressure

M = 0

Center of Pressure

Aerodynamic ForceLift

Drag

Moment = 0

V

+

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

Aerodynamic Center

V

+

Macx

y

Aerodynamic Center: The point on the airfoil where the moment is independent of angle of attack. Fixed for subsonic flight c/4. Fixed for supersonic flight c/2.

The moment has a nonzero value for cambered airfoils (negative for positively cambered airfoils). Moment is zero for symmetric airfoils.

Aerodynamic Force

Lift:

Force and Moment Coefficients

Drag: Moment:

Note: nondimensional coefficients!

Coefficients for NACA airfoils are found from charts in the Supplemental Data package or Appendix B of text.

Aerodynamic Force = FAERO = f( , , a , , V , S )

Using Dimensional Analysis FAERO = cf • q • S

Where: Cf = f (, Rec, Mach)

qScl l qScd d qScm m c

Lift-Curve Slope Terminology

[1] l=0 – Angle of attack () where the lift coefficient (cl) = 0, no lift is produced;

l=0 = 0 for a symmetric airfoil; l=0 < 0 for a positively cambered airfoil

Sample NACA Data

[2] cl– Lift-curve slope (dcl /d); ‘rise’ of cl over ‘run’ of for a linear

portion of the plot; 0.11/deg for a thin airfoil

[3] clmax– Maximum cl the airfoil can produce prior to stall

[4] stall – Stall angle of attack; at clmax; maximum prior to stall

cl

1

3

4

2

cl

Changes to lift and drag curves due to Reynolds number

At higher Reynolds numbers the boundary layer transitions to turbulent earlier, so it is more resistant to separation. Delayedseparation causes delayed stall and reduced pressure drag.

cd

cl

Low Re

cl

High Re

Changes to Lift Curves

1. Camber Positive camber

clcl

Zero Camber (symmetric)

Negative camber

Changes to Lift Curves

Without flaps

With flaps

2. Flaps

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

Without BLC

With BLC

cl

cl

DATA SHOWN ON NACA CHARTS (2421)

Airfoil Shape

Data point symbols for various Reynolds numbers (R)

Location of aerodynamic center (a.c.)

DATA SHOWN ON NACA CHARTS (2421)

Lift Curve :cl plotted against

DATA SHOWN ON NACA CHARTS (2421)

Drag Polar: cd plotted against cl

DATA SHOWN ON NACA CHARTS (2421)

Pitching moment coefficient at the quarter-chord point (cmc/4

) plotted against

DATA SHOWN ON NACA CHARTS (2421)

Pitching moment coefficient at the aerodynamic center (cmac

) plotted against cl

Example Problem

GIVEN: FIND: cl =

NACA 2421 airfoil cl = (cl / ) =

Reynolds number = 5.9x106 cd =

Angle of attack = 12° cmc/4 =

cma.c.=

clmax =

stall =

l=0 =

Example Problem (NACA 2421)

Reynolds Number

cl 1.3

Cm c/4 -0.025

Cl = (.78-.20)/(6°-0°) = 0.10/deg

Example Problem (NACA 2421)

Reynolds Number

cl 1.3

cd 0.017

Cm a.c. -0.045

Example Problem (NACA 2421)

Reynolds Number

clmax 1.38

stall 15°

l=0 -2°

Example ProblemWe just found: cl = 1.3 ; cd = 0.017 ; cmac

= -0.045 

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 V

2 = ½(0.002377 slug/ft3)(100 ft/sec)2 = 11.885 lb/ft2

 If c = 4 ft and S = 200 ft2, then: 

l = cl q S = ( ) (11.885 lb/ft2) (200 ft2) = lb

d = cd q S = ( ) (11.885 lb/ft2) (200 ft2) = lb

mac = cmac q S c

= ( ) (11.885 lb/ft2) (200 ft2) (4 ft) = ft-lb

1.3 3090.140.409

-0.045 -427.86

0.017

Next Lesson (T14)… Don’t come to the classroom – go

straight to the Aero Lab Prior to class

Read lab handout! In Class

Gather wind tunnel data