Supersonic Thin Airfoil Theory Andrew Ning In class we showed that - - PDF document

SLIDE 1

Supersonic Thin Airfoil Theory

Andrew Ning

In class we showed that the local pressure coefficient is given by (using small disturbance assumptions): Cp = 2θ

• M 2

∞ − 1

(1) where θ is positive when inclined into the freestream and negative when inclined away from the freestream. I won’t go through that derivation here as it was discussed in class. The part that we glossed over was how to get lift and drag coefficients from that result. As typically done in thin airfoil theory, we will separate an airfoil into a thickness distribution, a camber distribution, and an angle of attack. Specifically, we define the upper and lower surfaces as a superposition of camber and thickness distribution as follows: yu(x) = yc(x) + yt(x) (2) yl(x) = yc(x) − yt(x) (3) Note that yt is not the local thickness τ (it is 1/2 of that). It will be more convenient in the derivation to use this form. Using the formula for the local pressure coefficient: Cpu = 2

• M 2

∞ − 1

• −α + dyu

dx

• (4)

Cpl = 2

• M 2

∞ − 1

• α − dyl

dx

• (5)

The negative sign results from the way that θ is defined. Substituting in the camber and thickness distribu- tions: Cpu = 2

• M 2

∞ − 1

• −α + dyc

dx + dyt dx

• (6)

Cpl = 2

• M 2

∞ − 1

• α − dyc

dx + dyt dx

• (7)

The definition of the (inviscid) normal force coefficient is: cn = 1 c c (Cpl − Cpu)dx (8) Substituting in the result from above : cn = 2

• M 2

∞ − 1

1 c c (2α − 2dyc dx )dx (9) = 2

• M 2

∞ − 1

1 c

• 2α

c dx − 2 c dyc dx dx

• (10)

= 2

• M 2

∞ − 1

1 c (2αc − 2 yc|c

0)

(11) = 4α

• M 2

∞ − 1

(12) 1

SLIDE 2

The definition of the (inviscid) axial force coefficient is: ca = 1 c c

• Cpu

dyu dx − Cpl dyl dx

• dx

(13) Substituting in: ca = 1 c c

• 2
• M 2

∞ − 1

• −α + dyc

dx + dyt dx dyc dx + dyt dx

• −

2

• M 2

∞ − 1

• α − dyc

dx + dyt dx dyc dx − dyt dx

• dx

(14) Several terms appear in both expressions and so cancel out. Removing those terms leaves us with: ca = 2

• M 2

∞ − 1

1 c c

• −2αdyc

dx + 2 dyc dx 2 + 2 dyt dx 2 dx (15) For the first term under the integral, α is a constant and can be taken out. We already saw that c dyc dx dx = 0 (16) and so that whole term is zero. We are left with: ca = 4

• M 2

∞ − 1

dyc dx 2 + dyt dx 2 (17) where we define ζ = 1 c c ζ(x)dx (18) as a shorthand for convenience. Finally, lift and drag are related to the normal and axial forces as follows: cl = cn cos α − ca sin α (19) cd = cn sin α + ca cos α (20) Using a small angle approximation, consistent with thin airfoil theory, yields: cl ≈ cn − caα (21) cd ≈ cnα + ca (22) Conventionally, the caα term is neglected in the lift coefficient because it is of much smaller magnitude than cn (ca is small and α is small so their product is very small). However, in the drag calculation both cnα and ca are of similar magnitude. cl ≈ cn (23) cd ≈ cnα + ca (24) Thus: cl = 4α

• M 2

∞ − 1

(25) cd = 4

• M 2

∞ − 1

• α2 +

dyc dx 2 + dyt dx 2 (26) The (inviscid) pitching moment coefficient is: cmle = 1 c2 c

• Cpu − Cpl
• x + Cpu

dyu dx yu − Cpl dyl dx yl

• dx

(27) 2

SLIDE 3

Making substitutions: cmle = 1 c2 2

• M 2

∞ − 1

c

• −2α + 2dyc

dx

• x +
• −α + dyc

dx + dyt dx dyc dx + dyt dx

• (yc + yt)−
• α − dyc

dx + dyt dx dyc dx − dyt dx

• (yc − yt)
• dx

(28) Removing terms that cancel: cmle = 1 c2 2

• M 2

∞ − 1

c

• −2αx + 2dyc

dx x + yc

• −2αdyc

dx + 2dyc dx

2

+ 2dyt dx

2

+ yt

• −2αdyt

dx + 4dyc dx dyt dx

• dx

(29) Neglecting all second order terms. cmle ≈ 1 c2 2

• M 2

∞ − 1

c

• −2αx + 2dyc

dx x

• dx

(30) = 1 c2 2

• M 2

∞ − 1

• −2αc2

2 + c 2dyc dx xdx

• (31)

The second term we can integrate by parts: cmle = 1 c2 2

• M 2

∞ − 1

• −αc2 + 2(xyc)c

0 −

c ycdx

• (32)

The second term is zero, leading to cmle = − 2

• M 2

∞ − 1

• α + 1

c2 c ycdx

• (33)

= − 2

• M 2

∞ − 1

• α + yc

c

• (34)

3