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ME 408 Project#2 Lift Coefficient of Joukowski Airfoil (Due Wednesday 2/7/07) 1. Joukowski Transformation: An airfoil shape can be created by mapping a circle in the (x1, y1) plane to a new (x2, y2) plane with the transformation: 1 z2 z1 z1 z1 x1 1 y1 x2 1 y2 z2 1 is the imaginary are complex variables, and number. Also, the x-coordinate of the center of the circle must be negative (x0 < 0), its y-coordinate must be positive (y0 > 0), and the circle must pass through the point (1, 0) on the x1-axis. Hence the radius of the circle is: a 1 x0 2 y02 Hence the coordinates of the circle for each angular location (0 < < 2) are: a cos x0 x1 y1 a sin y0 (a) Show that the coordinates of the resulting airfoil shape are: x2 2 2 x1 x 1 y1 1 y2 x1 2 y1 2 2 2 y1 x1 y 1 1 x1 2 y1 2 (b) Select a center location (x0, y0) for the circle, and calculate the radius and the coordinates of the circle for each . Calculate the resulting Joukowski airfoil coordinates, and plot both the circle and the airfoil shape. Use the coordinates on the x2-axis to determine the chord length c. Notice that the point (1, 0) in the (x1, y1) plane is mapped to the sharp trailing edge of the airfoil. Joukowski Transformation 1.5 1.0 0.5 0.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 -0.5 -1.0 0.5 1.0 1.5 2.0 2.5 2. Kutta Condition and Lift Coefficient at =0 (a) Recall that at zero angle of attack (=0) and with no circulation (=0), the point at = 0 on the circle is a stagnation point, and the streamline from this stagnation point is simply the horizontal line y1 = y0. Map this streamline onto the (x2, y2) plane of the airfoil. Notice that the streamline along the lower surface of the airfoil has to make a 180-degree sharp turn to the stagnation point on the upper surface. (b) Since it is physically impossible for a streamline to change direction at the sharp trailing edge, the flow will add to itself a clockwise circulation to move the stagnation point on the circle clockwise to the point (1, 0), that is, through an angle: y0 sin 1 a This requirement is known as Kutta condition, which ensures the stagnation point on the airfoil to locate at the sharp trailing edge. Show that the circulation strength required is: 4 a U sin 4 U y0 where U is the free stream velocity. Since the stream function is 0 at the stagnation point on a circle, the streamline shape from the trailing stagnation point in polar coordinate is simply: U r 1 2 a r 2 sin a ln 2 r Trace this streamline from r = a to r = 3 by computing the corresponding (x1, y1) coordinates. Map the stagnation streamline onto the (x2, y2) plane, and verify that it indeed starts from the sharp trailing edge. (c) As we conclude from the study of potential flow past a circle, this circulation will generate a lift on the circle and therefore also on the airfoil: L U where is the fluid density. Compute the lift coefficient of the airfoil defined as: CL 2L U 2c 3. Lift Coefficient at Non-Zero Angle of Attack (> 0) For > 0, the stagnation point on the circle simply has to shift (clockwise to satisfy Kutta condition. Hence the clockwise circulation required is: 4aU sin (a) The lift is defined as the net force normal to the free stream. Show that the lift coefficient is: CL a 8 sin c (b) Compute and plot the lift coefficient from (the zero-lift angle of attack) to 100.