aero engineering 315 lesson 13 airfoils part ii. where does the moment come from? p u = upper...
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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