Aerodynamics 2

MEC 3706/MEC 3220

Assignment 1

Due Date: 14-1-2010 (11:30 am)

Q1.

Plot the pressure distribution over the symmetrical double wedge, 8 per cent thick supersonic aerofoil shown in figure below when the Mach number meets the upper surface (a) tangentially (b) 12º and (c) .9º. Compare the lift, drag and pitching moment coefficients for these incidents. Assume ε0=6º

Q2

A thin wing can be modeled as a 2m wide flat plate set at an angle of 5º to the upstream flow. If the wing is placed in a flow with a Mach number of 3.5 and a static pressure of 75 kPa, find using linearized theory the pressure on the upper and lower surfaces of the airfoil, lift and drag per span.

(Hint:

2

2∞

∞

∞ q = p M

γ

, γ=1.4)

2

Q3.

a- A thin symmetrical supersonic airfoil has parabolic upper and lower surfaces with a maximum thickness occurring at midchord. Using linearized theory, compute the lift to drag ratio when it is set at an angle of attack of 5º.

Figure 5

[Hint: y = A+ B (αx) + C (αx)2, and

dx

ε = dy ]

Q4.

Estimate the critical Mach number for NACA 0012 airfoil at 10º angle of attack. The pressure coefficient distribution over this airfoil, measured in wind tunnel at low speed is given in the following figure. Compute and draw the pressure distribution curve at Mach number of 0.75.

0 0.25 0.5 0.75 1

X/c

-0.5

0

0.5

1

1.5

2

2.5

3

Cp

3

Q1.

1- = ∫ ( − ) c

L P PU C c C C dx 0 L

for given figure, we obtain

= ∫ − + ∫ ( − ) / 2

0

/ 2

3 1 0 4 2 ( ) c c

L P P p P C c C C dx C C dx

But from linearized theory, we know

1

2

2 −

=

M

Cp

ε

then

( ) ( )⎟⎠

⎞

⎜⎝

⎛ − + −

−

= 2 2 3 1 2 4 2

1

1

2 c ε ε c ε ε

M c

CL

1- α = 6

6

12 12 6

6

M=3

c/2 c/2

6 6 0 0 1 ε = − = = radian

6 6 12 0.20933 2 ε = − − = − = − radian

6 6 12 0.20933 3 ε = + = = radian

6 6 1 0.0 4 ε = − + = = radian

( ) (0 ( 0.20933)) 0.148

2

0.20933 (0) 1

2

1

1

2

2

= ⎟⎠

⎞

⎜⎝

⎛ − + − −

−

=

M

CL

0.148

3 1

180

4 6

0.148

2

=

−

⎟ ⎟⎠

⎞

⎜ ⎜⎝

⎛

⎟⎠

⎞

⎜⎝

⎛

= =

π

L C

4

i- Wave drag

( ) ⎟⎠

⎞

⎜⎝

+ ⎛

−

= ∫ c

d x

M

C D L U

1

0

2 2

2 1

2 ε ε

( ) ( ) ⎟⎠

⎞

⎜⎝

⎛ − + + − +

−

= ∫ ∫ 1/ 2

0

1/ 2

0

2 2 2 2

2

0) (0.20933) ( ) ( 0.20933) (0) ( )

1

2

c

d x

c

d x

M

CD

= 0.0312 D C

iii- ∫ ( − )

−

−

= c

M L U xdx

c M

C

2 2 1 0

2 ε ε

⎟ ⎟⎠ ⎞

⎜ ⎜⎝

⎛

× + × −

−

−

=

⎟ ⎟

⎠

⎞

⎜ ⎜

⎝

⎛

+

−

−

= )

2 8

0.20933 (

8

0.20933

1

2

2

0.20933

2

1 0.20933

1

2 2 2 2

2 2

/ 2

/ 2 2

0

2

2 2

c c c

c M

x x

M c

C

c

c

c

M

0.20933 0.074

8

4

2.828

2 = − ⎟⎠

⎞

⎜⎝

⎛ ×

−

= M C

In similar manner we can obtain the CL, CD and CM for α=12º, and 9º 2- α=12º

CL=0.296, CD=0.077, CM= -0.148

3- α=9º

CL=0.222, CD=0.050, CM= -0.111

Q2.

Angle of attack that the upper surface makes to the flow is -5º=-0.0873 radians 0.0521

3.5 1

2 0.0873

2

= −

−

− ×

=

pupper C

Also

= −0.0521

−

=

∞

∞

q

p p

C upper

pupper

Given

2

2∞

∞

∞ = q p M

γ

then

5

75

2

0.0521 0.0521 0.0521 1.4 75 3.5

2

+

× ×

= − → = − × + = − ×

−

= ∞ ∞

∞

∞ p q p

q

p p

C upper

upper

pupper

p kPa upper = 41.5

For lower surface, similar method will be follow

Angle of attack that the lower surface makes to the flow is -5º=0.0873 radians 0.0521

3.5 1

2 0.0873

2

=

−

×

=

pupper C

= 0.0521

−

=

∞

∞

q

p p

C lower

pupper

Given

2

2∞

∞

∞ = q p M

γ

then

75

2

0.0521 0.0521 0.0521 1.4 75 3.5

2

+

× ×

= → = × + = ×

−

= ∞ ∞

∞

∞ p q p

q

C p p lower

lower

pupper

p kPa lower = 108.51

The lift will be given as

L ( p p )Acosα lower upper = −

Since α is small Cosα=1 and Sin α=α

Then

L = (108.51− 41.5)× 2× 2×1 = 268kN

D = (108.51− 41.5)× 2× 2× 0.0873 = 23.5kN

11.4

23.5

= 268 =

D

L

Q3.

The shape of the upper and lower surfaces is described by an equation that has the form y = A + Bαx + C(αx)2

For the upper surfaces this equation must satisfy the following conditions 6

B.C: , 0

2 2

x = 0→ y = 0, x = c → y = t x = c →...