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34
Mn2
sin( )
−
06099
37 520
.
sin( . )−
M2 is the Mach number ahead of the normal shock on the Pitot tube.
p
p
pp
o
o
o
3
2
1
1
1
ft 2
4.15 Shock polars do not give the solution directly.
35
4.16 The geometry of the airfoil is given below:
For face (a): When M1 = 2.0 and = 10, = 39.2
p
o3
37
Lift is the component of the total aerodynamic force in the y-direction:
L(per unit span) =
[(p4-p3) cos 20 + (p5-p2) cos 10]
4.17
38
p
2
T
2
3 M3
p
p
o3
3
p
atm
3
( )
T
T
o3
3
T
K
3
( )
0 49.76 3 36.73 1 2.8 270
5 54.76 3.27 54.78 0.670 3.139 241
(p2-p3) L D L/D
N/m2 N/m N/m
0 0 0 0 -
39
4.18 From Example 1.8, using the same flat plate with a chord of 3 feet, we have due to the
shear stress the force tangential to the flat plate (acting on one side) equal to 39.13 lb. Hence, the
skin friction drag due to the shear stress acting on both sides is
41
4.19
D = 2
c
cos 20
p2 sin 20 - p 2 c Tan 20 = 2 c (p2 - p) Tan 20
cd =
D
q S
D
p M c
' '
=
2
2
14 4 2
( . )( )
4.20 M2 – 1 = (3)2 – 1 = 8
1 +
−1
2
M2 = 1 +
12 1
2
.−
(3)2 = 1.9
−1
12 1
.+
43
tan = 0.715
= 35.57 For = 1.2
As a check on this answer, consider Eq. (4.17)
M
1
22
1
sin
−
4.21 From Eq. (4.9)
p
p
2
1
= 1 +
2
1
+
(
Mn1
2 – 1)
p
1
12 1
.+
44
p
2
