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REVIEW OF BASIC AERODYNAMIC PRINCIPLES 63
Solution. The Reynolds numberis given by
VL
Re = -
1)
50*3
= ~406~x~0 5
- 10.66 x l05
Now let us consider the two cases as follows.
Laminar firOw: from Eq. (1.13), we have
- 0.001286
Df = 7\p\/2SCf
= g *1.225 * 50z * 45 * 0.001286
- 88.6295 N
Turbulentltow: from Eq. (1.14), we have
. 0.455
Cf ='( o9~o R~258
0.455
N
= ( o9~o~0 66 x~05~2 58
-. 0.004418
N
Df = 2lp V2SCf
= ~: *1.225 * 502 * 45 * 0.004418
- 304.4829 N
Example 1.5
For a double wedge airfoil of 2 m chord and 4% thickness ratio, determine the
lift and wave-drag coefficients per unit span at a Mach number of 2.0 and a -
5 deg.
Cf = :k328
1.328
= .~~j R~_T~
:!
$
i
li
a4
. I:g
~
li
.~
:a
~
64 PERFORMANCE, STABILtTY, DYNAMICS, AND CONTROL
5
a = 5 deg = 575 3 = 0.08726 rad
4a
Cl = ./ff/-l
4 * 0.08726
= ~--I-
- 0.2015
4
Cdw = j~--f [-2+(-)2]
- 0.02128
Example 1.6
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