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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~

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 i
 li
a4
 . I:g
 ~
 li
 .~
:a
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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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