solution4

solution4 - 650:460 Aerodynamics Homework Assignment No. 4...

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Unformatted text preview: 650:460 Aerodynamics Homework Assignment No. 4 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 6.1 Show that the vorticity distribution γ ( θ ) = 2 α U ∞ 1 + cos θ sin θ satisfies the condition that the flow is parallel to the surface. Show that the Kutta condition is satisfied. Sketch the 2 γ/ U ∞ distribution as a function of x / c for a lift coefficient of 0.5. What is the physical significance of 2 γ/ U ∞ ? What angle of attack is required for a symmetric airfoil to develop a lift coefficient of 0.5 ? Solution Substituting the vorticity distribution into Eq (6.8), 1 2 π integraldisplay π o parenleftbigg 2 α U ∞ 1 + cos θ sin θ parenrightbigg sin θ cos θ − cos θ o d θ = U ∞ α 2 Problem 6.1 Thus 1 π integraldisplay π o 1 + cos θ cos θ − cos θ o d θ = 1 Using Eq (6.10) integraldisplay π o cos n θ d θ cos θ − cos θ o = π sin n θ o sin θ o the integral on the left equals 1. Thus, the equation is satisfied. 3 Problem 6.1 The Kutta condition requires lim θ → π γ ( θ ) = 0 Using L’Hospital’s rule lim θ → π γ ( θ ) = 2 α U ∞ lim θ → π − sin θ cos θ = 0 Thus, the Kutta condition is satisfied. The lift per unit depth is l = integraldisplay c o ρ ∞ U ∞ γ dx and thus ρ ∞ U ∞ γ represents the lift per infinitesimal chordwise distance per unit depth. 4 Problem 6.1 Since the lift is due to the pressure difference ρ ∞ U ∞ γ = p l − p u where l and u denote lower and upper surfaces. Equivalently ρ ∞ U ∞ γ = ( p l − p ∞ ) − ( p u − p ∞ ) and dividing by the dynamic pressure ρ ∞ U ∞ γ 1 2 ρ ∞ U 2 ∞ = c p l − c p u and thus 2 γ U ∞ = c p l − c p u 5 Problem 6.1Problem 6....
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This note was uploaded on 12/15/2009 for the course 650 460 taught by Professor Knight during the Spring '09 term at Rutgers.

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solution4 - 650:460 Aerodynamics Homework Assignment No. 4...

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