Unformatted text preview: n the
upper (u) and lower (l) surfaces are different but constant over each surface, i.e.,
and
, where
and
are constants and
> . The pressures on the
front (f) and rear (r) surfaces are different but constant over each surface, i.e.,
and
, where
and
are constants and
> . Shear stresses may be
ignored.
a. Calculate the lift force per unit span, .
b. Calculate the drag force per unit span, .
c. Calculate the moment per unit span about the leading edge,
, and comment
on its tendency to pitch the prism.
d. Determine the centre of pressure
. Neglecting shear stresses, we have:
∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ As > , ∫ ∫ ∫ ∫ is positive and thus it results in a pitchup moment. As the pressure distribution is uniform, it makes sense that center of pressure is at the midpoint. 3) A wing with a rectangular planform has a NACA 651210 aerofoil section (without
standard roughness). The aerofoil data are shown in the figure supplied with the exam.
The wing has a span of b = 5 m and a chord of c = 2 m. The span effectiveness and
Oswald efficiency factors are estimated to be e1 ≈ e ≈ 0.9. The wing is flying at
“standard” conditions (ρ∞ = 1.2 kg/m3, μ∞ = 1.8×105 Ns/m2, P∞ = 100 kPa, T∞ = 20°C) at
a speed of V∞ = 42 m/s and a geometric angle of attack of α = 8°.
a. Calculate the Reynolds number of the aerofoil section.
b. Calculate the lift curve slope for the wing section, a [in deg1 or rad1].
c. Calculate the lift coefficient of the wing, CL.
d. Calculate the total drag coefficient of the wing, CD.
e. Estimate the downwash velocity [in m/s] behind the wing.
We calculate directly from the formula as: From the graph given, we can see that the liftcurve slope We now use this new value of finite liftcurve slope to find the lift coefficient of the wing, CL
( ) From the graph given, we can see that the profile drag for given is To calculate the total drag coefficient, we add the profile drag with the induced drag: To calculate the downwash velocity, we use simple vectors as: ( ) ( ) (Problem 3 continued) ME 471.3
Introduction to Aerodynamics
Department of Mechanical Engineering
Winter 2014
Instructor – Rajat Chakravarty Midterm Examination Equation Sheet
TE TE LE LE N Pu cos u sin dsu A TE P sin u u cos dsu LE M LE u l l sin dsl l cos dsl TE P sin l LE TE P cos P cos u sin x Pu sin u cos y dsu LE TE P cos l l sin x Pl sin l cos y dsl LE c
c
1
C P,l C P,u dx c f ,u dyu c f ,l dyl dx c dx
dx 0
0
c
c dy
dy 1 c a C P ,u u C P ,l l dx c f ,u c f ,l dx c 0 dx
dx 0 c
c
dy
dy 1 c mLE 2 C P ,u C P ,l xdx c f ,u u c f ,l l xdx dx
dx c 0
0
c
c dy u
dy
1 2 C P ,u c f ,u y u dx C P ,l l c f ,l y l dx dx
dx
c 0 0 cl c n cos c a sin c d c n sin c a cos M
M
xcp LE LE
N
L
2
a0
CL
CL
C D cd i a
eAR
e1 AR a0 1 e AR 1 cn 2 0 2 0 u v y
x
x
y
1 1 2
2 r r r r r 2 2 z 2
1 1 vr v r r
r
r u
u
1 u z u
r r z
r z
z
r
2 1 ru u r r r z Additional Information
Rair = 287 J·kg1·K1
Sutherland Law (for viscosity of air):
3/ 2 T T0 S 5
2 T T S , where Sair ≈ 110.4 K, T0 = 273 K, μ0 = 1.71×10 Ns/m , and T is in [K]
0 0...
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This document was uploaded on 03/30/2014 for the course ME 471 at Maryland.
 Winter '14
 RajatChakravarty
 Mechanical Engineering

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