# α τ

```Implications of Linearized Supersonic Flow on Airfoil Lift &amp; Drag
To begin, we will divide the airfoil geometry into camber and thickness
distributions:
y
yc (x)
τ (x)
yu (x)
x
yl (x)
1
i
α∞
v
v
v
U ∞ ⇒ U ∞ = u ∞ cos α ∞ i + U ∞ sin α ∞ j
1
y u ( x) = y c ( x) + τ ( x)
2
1
y l ( x) = y c ( x) − τ ( x)
2
Note:
The x -axis, which is the axis defined through the leading edge and trailing edge
points, is used to define the freestream angle of attack. The chord line is the line
connecting i.e. to t.e. and, the chord length is the distance between these points.
To calculate the lift and drag, we need to integrate the pressure forces around
the airfoil.
v
v
v
F = −∫
pn ds
when n points out of the airfoil surface.
airfoil
Now, we begin specializing this formula for the assumptions of linearized flow,
i.e.
∗τ
∗
c
yc
&lt;&lt; 1
c
&lt;&lt; 1
∗ α ∞ &lt;&lt; 1
(small thickness)
(small camber)
(small α )
The normal on the upper surface is
Implications of Linearized Supersonic Flow on Airfoil Lift &amp; Drag
v
nu
y
v
tu
yu (x)
x
v dx v dy u v
j
tu =
i+
ds
ds
dy v dx v
v
j
nu = − u i +
ds
ds
dy v v
v
But, since we have thin airfoils ds ≅ dx , thus nu = − l i + 1 j
dx
v dy v v
Similarly, on the lower surface nl = l i − 1 j
dx
Thus, the force may be written:
v
c
c
v
v
F = − ∫ pu nu dx − ∫ p nl dx
0
0
l
v
c
c
⎛ dy v v ⎞
⎛ dy v v ⎞
F = − ∫ pu ⎜ − u i + j ⎟dx − ∫ pl ⎜ l i − j ⎟dy
0
0
⎝ dx
⎠
⎝ dx
⎠
v v c ⎛ dy
v c
dy ⎞
⇒ F = i ∫ ⎜ pu u − p l l ⎟dx + j ∫ ( p l − pu )dx
0
0
dx
dx ⎠
⎝
The drag component is in the freestream direction:
v ⎛ uv∞ ⎞ v
v
v
D ′ = F • ⎜⎜ ⎟⎟ = F • (cos α ∞ i + sin α ∞ j )
⎝ u∞ ⎠
1
∝∞
v v v v
D′ = F • i + F • j α ∞
⇒
c⎛
c
dy
dy ⎞
D ′ = ∫ ⎜ pu 0 − pl l ⎟dx + α ∞ ∫ ( pl − pu )dx
0
0
dx
dx ⎠
⎝
Similarly,
the lift is:
v
v
v
L ′ = F • (− sin α ∞ i + cos α ∞ j )
− ∝∞
v v
v v
= −F • i α ∞ + F • j
1
c⎛
c
dy
dy ⎞
L ′ = −α ∞ ∫ ⎜ pu u − pl l ⎟dx + ∫ ( pl − pu )dx
0
0
dx
dx ⎠
⎝
16.100 2002
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Implications of Linearized Supersonic Flow on Airfoil Lift &amp; Drag
Or, manipulating the L′ &amp; D′ slightly:
c⎡
⎛ dy
⎞
⎛ dy
⎞⎤
D ′ = ∫ ⎢ pu ⎜ u − α ∞ ⎟ − p l ⎜ l − α ∞ ⎟⎥dx
0
⎠
⎝ dx
⎠⎦
⎣ ⎝ dx
c⎡
dy ⎞
dy ⎞⎤
⎛
⎛
L ′ = ∫ ⎢ p l ⎜1 + α ∞ l ⎟ − pu ⎜1 + α ∞ u ⎟⎥ dx
0
dx ⎠
dx ⎠⎦
⎝
⎣ ⎝
p − p∞
, we can write this as:
1 2
ρu ∞
2
1 c ⎛ dy
⎛ dy
⎞⎤
⎞
c d = ∫ c pu ⎜ u − α ∞ ⎟ − c pl ⎜ l − α ∞ ⎟⎥ dx
0
c
⎝ dx
⎠⎦
⎝ dx
⎠
Then substituting in for the c p ≡
dy ⎞
dy ⎞⎤
1 c⎡ ⎛
⎛
c pl ⎜1 + α ∞ l ⎟ − c pu ⎜1 + α ∞ u ⎟⎥dx
⎢
∫
c 0⎣ ⎝
dx ⎠
dx ⎠⎦
⎝
As shown in Anderson
c p = 2θ
In supersonic linearized
M ∞2 − 1
flow where θ is the flow
⎛ dy u
⎞
direction relative to the
2⎜
−α∞ ⎟
dx
⎝
⎠
freestream (and
⇒ c pu =
M ∞2 − 1
assuming an isolated
airfoil).
⎛ dy
⎞
2⎜ − l + α ∞ ⎟
⎠
Upper surface
c pl = ⎝ dx
θu
M ∞2 − 1
dyu
cl =
α∞
dx
dy u
−α∞
dx
Lower surface
⇒ θu =
dx
α∞
− dyl
α∞
⇒ θl = −
θl
dy l
+ α∞
dx
Next, let’s substitute c pu and c pl into cl :
16.100 2002
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Implications of Linearized Supersonic Flow on Airfoil Lift &amp; Drag
cl =
dy ⎞
dy ⎞⎤
1 c
2
2
⎛ dy u
⎞⎛
⎛ dy l
⎞⎛
+ α ∞ ⎟⎜1 + α ∞ l ⎟ −
− α ∞ ⎟⎜1 + α ∞ u ⎟⎥ dx
⎜−
⎜
∫
0
c
dx ⎠
dx ⎠⎦
⎠⎝
⎠⎝
M ∞2 − 1 ⎝ dx
M ∞2 − 1 ⎝ dx
small
cl ≈
=
2
c M ∞2
small
⎛ dy l dy u
⎞
−
+ 2α ∞ ⎟dx
0
dx
dx
⎠
c
∫ ⎜−
−1 ⎝
c dy
⎡ c dy l
⎤
u
dx
dx + 2α ∞ c ⎥
−
−
∫
⎢ ∫0 dx
0 dx
2
⎦
c M ∞ −1 ⎣
2
dy l
dx = y l (c) − y l (0) = 0 − 0 = 0
0 dx
c dy
u
And, similarly, ∫
dx = 0
0 dx
But
⇒
∫
c
cl =
4α ∞
Important result!!
M ∞2 − 1
∗ cl is linear with α ∞ but note the slope is different than subsonic case
⎛ dc
⎞
2π
⎜ l =
⎟.
2 ⎟
⎜ dα
1− M∞ ⎠
⎝
∗ cl does not depend on camber! All the y -dependence has disappeared in this
result. Thus, cl also does not depend on thickness.
Now, let’s look at cd :
2
2
⎡⎛ dy u
⎞ ⎛ dy l
⎞ ⎤
cd =
∫ ⎢⎜ − α ∞ ⎟⎠ + ⎜⎝ dx − α ∞ ⎟⎠ ⎥⎥dx
c M ∞2 − 1 0 ⎢⎣⎝ dx
⎦
2
c
2
⎡⎛ dy u ⎞ 2
⎤
⎛ dy u dy l ⎞ ⎛ dy l ⎞
=
− 2α ∞ ⎜
+
+⎜
+ 2α ∞2 ⎥ dx
⎜
⎟
⎟
⎟
⎢
∫
dx ⎠ ⎝ dx ⎠
⎝ dx
c M ∞2 − 1 0 ⎢⎣⎝ dx ⎠
⎥⎦
2
c
This term will
integrate to 0
⎡⎛ dy u ⎞ 2 ⎛ dy l ⎞ 2
⎤
⇒ cd =
⎜
⎟ +⎜
⎟ + 2α ∞2 ⎥dx
⎢
∫
c M ∞2 − 1 0 ⎢⎣⎝ dx ⎠ ⎝ dx ⎠
⎥⎦
2
16.100 2002
c
4
Implications of Linearized Supersonic Flow on Airfoil Lift &amp; Drag
⇒
cd =
2
2
1 c ⎡⎛ dy u ⎞ ⎛ dy l ⎞ ⎤
+
∫ ⎢⎜ ⎟ + ⎜ ⎟ ⎥ dx ≥ 0
M ∞2 − 1
M ∞2 − 1 c 0 ⎢⎣⎝ dx ⎠ ⎝ dx ⎠ ⎥⎦
4α ∞2
2
Note: cd is &gt;0 unless ∝∞ = 0 and airfoil is a plate ( y u = y l = const ⇒ y u = y l = 0) .
A little manipulation gives another form dependent on the camber and thickness:
cd =
4α ∞2
2
1 c
1
1 c ⎛ dτ ⎞
(
)
+
dy
dx +
c
∫
∫ ⎜ ⎟ dx
0
2
2
2
M ∞ −1
M ∞ −1 c
M ∞ − 1 c 0 ⎝ dx ⎠
4
2
Thus, for a given cl , the lowest cd occurs when the airfoil is a flat plate
( yc = τ = 0) !!
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