Lesson 13 Airfoils Part II Aero Engineering 315 Lesson 13 Airfoils Part II

Where does the moment come from? Fnet M PL = Lower surface pressure distribution PU = Upper surface pressure distribution Note: M is negative for this example In general: M is < 0 for positive camber M is = 0 for symmetric airfoils M 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 Force Lift Moment = 0 + Drag 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 a y Aerodynamic Force Mac x + 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.

Force and Moment Coefficients Aerodynamic Force = FAERO = f( a , m , a , r , V , S ) Using Dimensional Analysis  FAERO = cf • q • S Where: Cf = f (a, Rec, Mach) c Lift: Drag: Moment: Note: nondimensional coefficients! Coefficients for NACA airfoils are found from charts in the Supplemental Data package or Appendix B of text.

Lift-Curve Slope Terminology 3 2 1 a Sample NACA Data 4 [1] al=0 – Angle of attack () where the lift coefficient (cl) = 0,  no lift is produced; l=0 = 0 for a symmetric airfoil; al=0 < 0 for a positively cambered airfoil [2] cla – 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

Changes to lift and drag curves due to Reynolds number High Re Changes to lift and drag curves due to Reynolds number c d l c l a Low Re At higher Reynolds numbers the boundary layer transitions to turbulent earlier, so it is more resistant to separation. Delayed separation causes delayed stall and reduced pressure drag.

Changes to Lift Curves cl 1. Camber a Positive camber Zero Camber (symmetric) cl Negative camber a

Changes to Lift Curves c c a a 2. Flaps Without flaps With flaps 3. Boundary Layer Control (BLC) or increasing Reynolds Number c l a Without BLC With BLC

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 cma.c.= clmax = stall = l=0 = GIVEN: FIND: cl = NACA 2421 airfoil cla = (cl / ) = Reynolds number = 5.9x106 cd = Angle of attack = 12° cmc/4 = cma.c.= clmax = stall = l=0 =

Example Problem (NACA 2421) cl  1.3 Cla = (.78-.20)/(6°-0°) = 0.10/deg Cm c/4  -0.025 Reynolds Number

Example Problem (NACA 2421) cd  0.017 cl  1.3 Cm a.c.  -0.045 Reynolds Number

Example Problem (NACA 2421) clmax  1.38 l=0  -2° stall  15° Reynolds Number

Example Problem We 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 r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 3090.1 0.017 40.409 -0.045 -427.86

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