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Unformatted text preview: 4.16) Using shockexpansion theory, calculate the lift and drag (in pounds) on a symmetrical diamond airfoil of semiangle 15 ε = (see Fig. 4.35) at an angle of attack to the free stream of 5. when the upstream Mach number and pressure are 2.0 and 2116 2 / lb ft , respectively. The maximum thickness of the airfoil is 0.5 t ft = . Assume a unit length of 1 ft in the span direction (perpendicular to the page in Fig. 4.35). Also calculate the down/up wash angle. First off, to better visualize this, a diagram is in need. For ease of referencing, I have broken the diamond into 6 regions. To start our problem, let us start with Regions 1 & 2 and progress down the body to Regions 3 & 4. Given 15 ε = and 5 α = , it is easy to see that our angle of inclination for Region 1 is defined as 10 ε α = . Given that 2.0 M b = and utilizing our M θ β relationship, we can state that: ( 29 ( 29 2 2 1 1 2 1 sin 1 tan 2cot cos2 2 M M β ε α β γ β = + + Utilizing the fancy MATLAB iterative solver it can determined that, 1 39.3139 β = . This allows us to determine our Mach number normal to the shock wave as, ( 29 1 1 sin 2.0sin 39.3139 1.2671 n M M β = = = From here we can calculate the normal component of the Mach number downstream of the oblique shock and the tangential component of the Mach number with respect to the diamond surface as, ( 29 ( 29 ( 29 ( 29 ( 29 1 1 2 2 2 1 2 2 1.4 1 1 1 1.2671 1 2 2 0.6451 1 1.4 1 1.4 1.2671 2 2 n n n M M M γ γ γ + + = = = 1 0.8032 n M = [ ] ( 29 ( 29 1 1 1 0.8032 1.6405 sin 39.3139 10 sin n M M β ε α = = = From this point our flow encounters an expansion angle. To calculate the Mach number downstream of our expansion, lets obtain the PrandtlMeyer Function value. This value is defined as, ( 29 ( 29 1 2 1 2 1 1 1 1 1 1 tan 1 tan 1 1 1 v M M M γ γ γ γ + = + Utilizing a MATLAB code, we can state that, ( 29 1 1 16.0575 v M = . We also know that, 3 1 1 v v λ = + , where we can define our expansion angle as the total 180 degree expansion minus the interior angle of the diamond, i.e. 1 180 λ θ = . From geometric arguments, we can state that theta is defined as, 360 4 180 2 150 2 ε θ ε = = = A , (360 degrees in a square, minus our known angles yields the sum of our two theta angles). Utilizing this value of Theta, we can state that,...
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This note was uploaded on 05/05/2008 for the course MAE 362 taught by Professor Staff during the Spring '07 term at ASU.
 Spring '07
 Staff

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