曝光臺 注意防騙 網曝天貓店富美金盛家居專營店坑蒙拐騙欺詐消費者

coefficient about the leading edge at zero-lift was found to be equal to -0.02. At
u = 8 deg, CL  - 0.7, Cd - 0.04, and Cm.,t - -0.20, determine the location of the
aerodynamic center.
Solution.   From Eq. (1.37),
Cm = Cmac - Xoc CL
We have Crnac - Cmo -. - 0.02, and Ci - 0.7, Cm = -0,2 at a - 8 deg. Then,
                                            Cmo - Cm
                           xac - -
                  Cl
                                               -0.02 - (-0.20)
                                                                                             =                     0-. 7
                                    - 0.2571
Thus, the aerodynamic center is located at 25.71eVo chord from Lhe leading-edge.
Example 1.3
      Deternune the mean aerodynamic chord and its spanwise location for an aircraft
wing having a root chord of 6.1 m, tip chord of 3.28 m, leading-edge sweep of
30 deg, and a semispan of 15.25 m.
So/ution.   From Eqs. (1.25) and (1.26), we have
- = 24,r(::-+A) )
                        )
    ymac =  :(L++2A} )
We have Cr - 6.1, b - 2 * 15.25 - 30.5, and the taper ratio A = 3.28/6.1  -
0.5377. Substituting, we get
  c - 4.8313 m
yIWK = 6.8608 m
  Example 1.4
  For a straight rectangular wing having a chord of 3 m and a span of 15 m,
determine the skin-friction coefficient and slan-friction drag at a velocity of 50
m/s, assuming the boundary layer to be laminar and the boundary layer to'9e fully
turbulent. Assume p - 1.225 kg/m3 and v = 1.4607 x 10-5 rri2/s.
::!
  3
:i
 /
;/
  ::
   :.
  ..
.y
":/
J.
?,
/f
 ' :i
::::-::
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
　
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