ME 471 - Midterm Exam 1 Solutions

# A schematic is shown in the figure below the

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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 pitch-up 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 651-210 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∞ = 1.8×10-5 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 deg-1 or rad-1]. 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 lift-curve slope We now use this new value of finite lift-curve 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·kg-1·K-1 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.

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