0.40 m 0.21 m 0.02 m NACA Airfoils 6-Feb-08 AE 315 Lesson 10: Airfoil nomenclature and properties 1 Definitions: Airfoil Geometry z Mean camber line Chord line x Chord x=0 x=c Leading edge Trailing edge NACA Nomenclature NACA 2421 1 st Digit: Maximum camber is 2% of 2D airfoil chord length, c (or 3D wing mean chord length, c). 2 nd Digit: Location of maximum camber is at 4/10ths (or 40%) of the chord line, from the LE. 3 rd & 4 th Digits: Maximum thickness is 21% of c (or c). c = 1 m
Lift-Curve Slope Terminology 3 2 1 4 Sample NACA Data [1] l=0 Angle of attack () where the lift coefficient ( ) = 0, no lift is produced; l=0 = 0 for a symmetric airfoil; l=0 < 0 for a positively cambered airfoil. [2] Lift-curve slope (d /d); rise of over run of for a linear portion of the plot; 0.11/deg for a 2D (thin) airfoil (sometimes called a 0 ) [3] max Maximum the airfoil can produce prior to stall [4] stall Stall angle of attack; at max ; maximum prior to stall 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 1 st Digit: Maximum camber is _% of 2D airfoil chord length, c 2 nd Digit: Location of maximum camber is at /10 ths of the chord line, from the LE. 3 rd & 4 th NACA 6716 Digits: 0.14 Maximum thickness is _% of c). 0.09 y/c 0.04-0.01-0.06 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x/c
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 Span (b) Chord (c) Planform Area (S) S = b * c Angle of Attack, Airfoil Forces and Moments Lift Moment + Aerodynamic Force V Relative wind Drag 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 and pressure forces Lift Total Aerodynamic Force (Sum of Pressure and Shear) Pressure V free stream Drag Figure 3.10, Page 75 Shear V Center of Pressure Lift Aerodynamic Force 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 ) Increasing angle of attack causes the center of pressure to move forward while decreasing angle of attack moves the center of pressure backwards. y Aerodynamic Center x M ac + V 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 Lift: Drag: Moment: l cl q S c c d m d q S m q Sc Note: nondimensional coefficients! 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 : plotted against
Drag Polar: c d plotted against Pitching moment coefficient at the quarter-chord point (c mc/4 ) plotted against Pitching moment coefficient at the aerodynamic center (c mac ) plotted against
Daily quiz Given: NACA 2421 Airfoil Reynolds Number = 5.9x10 6 Angle of Attack = 12 Find: = = ( / ) = c d = c m c/4 = c m a.c. = max = stall = l=0 = Example Problem (NACA 2421) 1.3 = 0.6/6 = 0.1/ C m c/4-0.025 Reynolds Number Example Problem (NACA 2421) c d 0.018 1.3 C m a.c. -0.04 Reynolds Number
Example Problem (NACA 2421) max 1.4 l=0-2 stall 15 Reynolds Number Example Problem We just found: = 1.3 ; c d = 0.018 ; c mac = -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/ft 3 ) (100 ft/sec) 2 = 11.885 lb/ft 2 If c = 4 ft and S = 200 ft 2, l = q S = (1.3) (11.885 lb/ft 2 ) (200 ft 2 ) = 3090 lb d = c d q S = (0.018) (11.885 lb/ft 2 ) (200 ft 2 ) = 42.8 lb m ac = c mac q S c = (-0.04) (11.885 lb/ft 2 ) (200 ft 2 ) (4 ft) = - 380 ft-lb Changes to Lift Curves 1. Camber Positive camber Zero Camber (symmetric) 2. 2D vs 3D wings C L C L Airfoil (2-D) Wing (3-D) Negative camber c C = l L 57.3c 1+ l π ear C = C ( ) L L L = 0 6-Feb-08 AE 315 Lesson 10: Airfoil nomenclature and properties 24
Changes to Lift Curves 3. Flaps Without flaps With flaps 4. Boundary Layer Control (BLC) or increasing Reynolds Number Without BLC With BLC 6-Feb-08 AE 315 Lesson 10: Airfoil nomenclature and properties 25 3-D Effects on Lift and C L Airfoil C L Wing Notice the slope is decreased for the wing and the zero lift angle of attack is unchanged. 6-Feb-08 AE 315 Lesson 10: Airfoil nomenclature and properties 26