Download NACA Airfoils

NACA Airfoils
6-Feb-08
AE 315 Lesson 10: Airfoil nomenclature and properties
1
Definitions: Airfoil Geometry
z
Mean camber line
x
Chord line
Chord
x=0
Leading edge
x=c
Trailing edge
NACA Nomenclature
1st
NACA 2421
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).
0.40 m
0.21 m
0.02 m
c=1m
1
Lift-Curve Slope Terminology
cl c
l
3
2
1
α
4
Sample NACA Data
[1] αl=0 – Angle of attack (α) where the lift coefficient (cl) = 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 (dcl/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
y/c
0.09
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
2
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
an
Sp
)
(b
Planform Area (S)
S=b*c
Chord (c)
Angle of Attack,
Airfoil Forces and Moments
Lift
Aerodynamic Force
Moment
+
V∞
α
Drag
Relative
wind
Angle of attack (α) : 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
3
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∞
α
Drag
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 the center of pressure backwards.
Aerodynamic Center
y
x
V∞
Mac
+
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.
4
Force and Moment Coefficients
Lift:
Drag:
Moment:
l
q∞ S
d
cd ≡
q∞ S
cl ≡
cm ≡
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?
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 α
5
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 α
DATA SHOWN ON NACA CHARTS (2421)
Pitching moment coefficient at the aerodynamic
center (cmac) plotted against cl
6
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
Example Problem (NACA 2421)
cd ≈ 0.018
cl ≈ 1.3
Cm a.c. ≈ -0.04
Reynolds Number
7
Example Problem (NACA 2421)
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 ft2,
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
2. 2D vs 3D cl
CL
wings
cl α
CL α
6-Feb-08
Airfoil
(2-D)
Wing
(3-D)
α
C
C
Lα
Lα
=
1+
=C
AE 315 Lesson 10: Airfoil nomenclature and properties
c
lα
57 . 3 c
lα
π eAR
Lα
(α − α
L=0
)
24
8
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
AE 315 Lesson 10: Airfoil nomenclature and properties
25
3-D Effects on Lift
c l and CL
Airfoil
cl α
CL α
Wing
α
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
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