Solution Exercise Sheet 1 Mec 3707

Exercise Sheet 1: Incompressible Flow over Finite Wings MEC 3707 Aerodynamics 2 Semes. 2, 12/13 Quiz1 ( /3/2012, 2:50-3:15 pm) 1- Consider a tapered wing with an aspect ratio of 6.5, an induced drag factor δ= 0.045 and a zero lift angle of attack of -3o. At angle of attack of 4.4o, the induced drag coefficient for this wing is 0.015. Calculate the induced drag coefficient for a similar wing (a tapered wing with the same airfoil section) at the same angle of attack, but with an aspect ratio of 8.5. Assume that the induced factors for drag and the lift slop, δ and τ, respectively, are equal to each other (δ=τ). Also, for AR=8.5, δ=0.095. [Hint: e=0.85] 2- Consider a finite wing with an aspect of ratio of 7; the airfoil section of the wing is a symmetric airfoil with an infinite wing lift slope of 0.11 per degree. The lift to drag ratio for this wing is 29 when the lift coefficient is equal to 0.35. If the angle of attack remains the same and the aspect ratio is simply increased to 10 by adding extension to the span of the wing, what is the new value of the lift-to-drag ratio? Assume the span efficiency factors e=e1=0.9 for both cases.

3- Consider the Beechcraft Baron 58 aircraft cruising such that the wing is at a four degree angle of attack. The wing of this airplane has an NACA 23015 airfoil (αL=0=-1° and a0=0.113 per degree), aspect ratio of 7.61 and a taper ratio of 0.45. Calculate CL and CD for the wing. Assume τ=δ=0.01, αstall=12° and Re=8.9x105.

4- The Piper Cheroke ( a light, single engine general aviation aircraft) has a wing area of 170 ft2 and wing span of 32 ft. Its maximum gross weight is 2450 lb. The wing uses NACA 65-415 airfoil, which has a lift slope of 0.1033 degree-1 and αL=0=-3°. Assume τ=0.12. If the airplane is cruising at 120 mi/h at standard sea level at its maximum cross weight and is in straight and level flight, calculate the geometric angle of attack of the wing.

Solution Q1. Ans. Given data - AR1=6.5, AR2=8.5, - Same airfoil section: α L = 0

= −3
=constant

α1 = α 2 = 4.4
- δ1 = 0.045 , δ 2 = 0.095 , τ 2 = δ 2 = 0.095 - C D ,1 = 0.015

-

Needed Data: C D , 2 = ? Solution steps -

Calculate CL1 For the wing with AR=6

From Eq. 5.61

C D ,i

πC AR C L2 = (1 + δ ) → CL2 = D,i πAR 1+ δ

→ CL2 =

π (0.015)(6.5) 1 + 0.045

→ CL = 0.541

Lift slop (a )for wing with AR=6 :

a=

dC L CL (α = 4.4
) − C L =0(α =−3
) 0.541 − 0 = = dα α L − α l =0 4.4
− (−3
)

→ a = 0.0731 / deg ree = 4.188 / rad From Eq. 5.70

dCL a0 =a= = 4.188 a0 dα 1+ ( )(1 + τ ) πAR

→ 4.188 =

a0 a0 =  1.045a0  1 + 0.0512a0 1 +    6.5π 

→ a0 = 5.33 / rad Since the airfoil section is the same for both wings, a0 is a constant * Lift slop for the wing with AR=8.5

5.33 5.33 1+ ( )(1 + 0.095) π (8.5) → a = 4.37 / rad = 0.0763 / deg ree

a=

•

Calculate CL For the wing with AR=8.5

(

)

CL = a(α − α l =0 ) = 0.0763 4.4
− (−3
) = 0.565 Induced drag for wing with AR=8.5

C L2 = (1 + δ ) = (0.565) (1.095) = 0.0131 πAR π (8.5) 2

C D ,i

Q2. First solve the angle of attack and the profile drag coefficient, which say the same in this problem deg/radian= 180/π= 57.3 a0 C  a  C L = aα = α → α = L 1 + 0  = a0 a 0  πeAR  1+ πe1 AR 0.35  0.11 × 57.3  1+ = 0.0732radian = 4.2
0.11 × 57.3  π × 0.9 × 7 

The profile drag can be obtained as follows CL 0.35 CD = = = 0.012 29  CL     CD 

C L2 C2 (0.35) = 0.0058 → c d = C D − L = 0.012 − πeAR πeAR π (0.9)(7 ) Increasing the aspect ratio at the same angle of attack increases CL and reduces CD. For AR=10, we have a0 0.11 × 4.2 = = 0.3778 C L = aα = a0 0.11 1+ 1+ π πe1 AR × 0.9 × 10 57.3 2

C D = cd +

C L2 0.3778 2 C D = cd + = 0.0058 + = 0.0062 + 0.005048 = 0.01085 πeAR π × 0.9 × 10 Hence, the new value of L/D is

C L 0.3778 = = 34.7 C D 0.0112

Q3. a0 = 0.113(57.3) = 6.47 per rad 1 1 e= = = 0.99 1 + δ 1 + 0.01 a0 6.47 Assumed τ = δ → a = = = 5.08 per rad=0.0887 per deg 1 + (0.271)(1.01)  a0  1+  (1 + τ )  πAR  Since stall angle of attack is given as 12 degree, at α=4°, the lift curve can be consider linear, then C L = a (α − α L =0 ) = 0.0887(4 − (− 1)) = 0.443 C L2 (3.1) πeAR Here, Cd, is the section drag coefficient given in given figure. Note that in the given figure Cd is plotted versus the section lift coefficient cl. To accurately read cd , we need to know the value of cl actually sensed by the airfoil section. The value of the airfoil cl at effective angle of attack, αeff. Since e=0.99, the elliptical lift distribution can be assumed C 0.443 = 0.0185rad = 1.06
αi = L = πAR π (7.61) α eff = α − α i = 4
− 1.06
= 2.94
≈ 3
C D = Cd +

The lift coefficient sensed by the airfoil is then cl = a 0 (α eff − α L =0 ) = 0.113(3 − (− 1)) = 0.452

From the given cl vs cd curve, for cl=0.542 , we have cd=0.0065 Return to Eq. 3.1 2 C L2 ( 0.443) C D = Cd + = 0.0065 + = 0.0148 πeAR π (0.99 )(7.61) Q4. AR =

b 2 32 2 = = 6.02 S 170

At standard sea level , ρ ∞ = 0.002377

slug ft 3

 88 ft / s  ft  = 176 V∞ = 120mph s  60mph  q∞ =

1 1 lb ρ ∞V∞2 = (0.002377 )(176)2 = 36.8 2 2 2 ft

a0 = 0.1033 / deg ree = 5.92 / rad CL =

L W 2450 = = = 0.3916 q ∞ S q ∞ S 36.8 × 170

a0

a= 1+

a0 (1 + τ ) πAR

=

5.92 = 4.38 / rad = 0.0764 / deg 5.92 (1 + 0.12) 1+ π (6.02 )

C L = a(α − α L =0 ) → α =

CL 0.3916 + α L=0 = − 3
= 2.12
a 0.0764