Carlos Nunes Hw5 Ae515
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Transcript of Carlos Nunes Hw5 Ae515
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AE 515 SPRING 2015
HW#5
CARLOS ALBERTO NUNES
UIUC
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2
21
pCM
=
(1.1)
And:
uu
ll
yxyx
=
= +
(1.2)
Using the given equations for the upper and lower surfaces:
max max
max
max
1 12 1 2
12 1
2 1 2
u
u
u
y t tx xcx c c c c c c
y t x xx c c c
y t xx c c
=
=
=
(1.3)
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Since l uy yx x
=
then we have:
max2 1 2ly t xx c c
= (1.4)
Plug back eq. 1.3 and 1.4 in eq. 1.1 to find the equation for the pressure distribution:
max2
max2
2 2 1 21
2 2 1 21
u
l
p
p
t xCc cM
t xCc cM
=
= +
(1.5)
For = 0 we found the following distribution, where both up
C and lp
C are the same
When = 6:
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We know that:
2
41
LCM
=
(1.6)
The drag coefficient is given by:
( )2
2 2
2 2
4 21 1
D u lCM M
= + +
(1.7)
Where:
2
2
0
1 c uu
y dxc x
=
(1.8)
For this geometry 2 2u l = , then substituting eq. 1.4 in eq. 1.8 and integrating we get:
2
2 max43u
tc
=
(1.9)
Therefore, the eq. 1.7 may be re-written as:
22
max2 2
4 2 831 1
DtCcM M
= + (1.10)
Using eq. 1.6 and 1.10 we found the following values for the two given angle of attack:
= 0 = 6 CL 0 0.3700 CD 2.33E-02 6.25E-02
To calculate the center of pressure location we did the following:
I. Calculate the lift per unit length of each element of upper and lower surface:
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'
' 4
cos
cos
upper upper
lower upper
upperi p
loweri p
yL C q x
x
yL C q xx
=
=
(1.11)
II. Compute the total lift per unit length of each element:
( )' ' 'upper loweri i itotal
L L L= + (1.12)
III. Determine the total lift and moment ( about the leading edge) of the airfoil:
' '
1
' '
1
wing i
wing i
n
in
ii
L L
M L x
=
=
=
=
(1.13)
IV. And finally find the pressure center location:
'
'wing
cpwing
Mx
L= (1.14)
L'/q_inf 0.3690 M'/q_inf -0.1845
xcp 0.500
The result is in agreement with the linear theory for supersonic airfoils, since it is at 50% of the
airfoil chord.
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As we can see, the 2nd order theory shows that we have less suction in the upper surface and more positive
pressure in the lower surface for positive angles of attack when compared with the linear theory.
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The wave drag is given by:
2
24
64W
VD ul
= (1.15)
Using the following values for the given altitude and speed:
We found:
123.30[ ]wD lbs=
-
The previous charts and table shows how the wave drag changes with the length for a fixed volume body.
We can see that the drag drastically decays as we increase the body length, but to calculate the body total
drag we have also to consider the skin friction drag, that is proportional to the wet surface area, and then to
the body length. Therefore, we must find an optimum length that minimizes the total drag, which for this
case must be something between 10 and 20 ft.
The wave drag coefficient for a von Karman body is given by:
2 21 0.04
5wb
DdCl
= = =
(1.16)
2
2
78.53
6142.77[ / ]BS ft
q lbs ft
=
=
-
von Karman body wave drag:
19298.09[ ]ww B D
D S q C lbs= = (1.17)
The conical forebody wave drag:
20262.99[ ]wD lbs= (1.18)
The radial distribution in the von Karman body is given by (as function of ):
( ) ( ) ( )2sin 2
2s l
r
= +
(1.19)
And for the conical body:
( )2
bdxr xl
= (1.20)
Then we have the following radius distribution along the body length: