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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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