solution11

# solution11 - 650:460 Aerodynamics Homework Assignment No 11...

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Unformatted text preview: 650:460 Aerodynamics Homework Assignment No. 11 Prof. Doyle Knight Tel: 732 445 4464, Email: [email protected] Office hours: Tues and Thur, 4:30 pm - 6:00 pm and by appointment Fall 2009 1 Problem 10.5 Repeat Problem 10.3 using the shock expansion technique. What is the maximum angle of attack at which this airfoil can be placed and still generate a weak shock wave ? Note: We will solve the problem for α = 2 . 29 ◦ only. 2 Problem 10.5 Region 2 The flow from Region 1 to Region 2 passes through a weak shock wave. The turning angle δ 2 = 4 . 58 ◦ which is the sum of the angle of attack and the half angle of the wedge. The shock angle is θ 2 = 33 . 9 ◦ and the pressure ratio p 2 / p 1 = 1 . 286. Using Fig. 8.12, c p 2 = ( p 2- p 1 ) 1 2 ρ 1 U 2 1 = ( p 2 / p 1- 1) 1 2 γ M 2 1 = 0 . 102 The Mach number in Region 2 is M 2 = 1 . 836 using Fig. 8-12 or Eq (8-75). 3 Problem 10.5 Region 4 The flow expands from Region 2 to Region 4. Using the left-running characteristic ν 2- θ 2 = ν 4- θ 4 From Table 8.2, ν 2 = 21 . 7 ◦ . Thus ν 4 = ν 2 + θ 4- θ 2 = 21 . 7 ◦ + 4 . 58 ◦ = 26 . 3 ◦ Thus, from Table 8.2, M 4 = 1 . 997. The total pressure is constant between Region 2 and Region 4, and thus p 4 p 2 = p 4 p t 4 p t 4 p t 2 p t 2 p 2 = p 4 p t 4 p t 2 p 2 = [1 + ( γ − 1) 2 M 2 2 ] γ/ ( γ − 1) [1 + ( γ − 1) 2 M 2 4 ] γ/ ( γ − 1) = 0 . 777 4 Problem 10.5 Thus p 4 p 1 = p 4 p 2 p 2 p 1 = 0 . 9945 and hence c p 4 = ( p 4- p 1 ) 1 2 ρ 2 1 U 2 1 = ( p 4 / p 1- 1) 1 2 γ...
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solution11 - 650:460 Aerodynamics Homework Assignment No 11...

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