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HW6 - 4.6 A supersonic stream at M 1 3.6 flows past a...

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4.6) A supersonic stream at 1 3.6 M = flows past a compression corner with a deflection angle of 20. . The incident shock wave is reflected from an opposite wall which is parallel to the upstream supersonic flow, as sketched in Fig. 4.18. Calculate the angle of the reflected shock relative to the straight wall. Given our handy M θ β - - relationship, ( 29 2 2 1 2 1 sin 1 tan 2cot cos2 2 M M β θ β γ β - = + + , we can develop an iterative solution with MATLAB. Utilizing the code, it can be determined that 34.1103 β = ( 29 1 1 sin 3.6sin 34.1103 2.0188 n M M β = = = ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 1 2 2 2 1 1.4 1 1 1 2.0188 1 2 2 0.3297 1 1.4 1 1.4 2.0188 2 2 n n n M M M γ γ γ - - + + = = = - - - - 2 0.5742 n M = ( 29 ( 29 2 2 0.5742 2.3553 sin sin 34.1103 20 n M M β θ = = = - - Referring once again to our M θ β - - relationship and our handy code, for 20 θ = and 2 2.3553 M = , then 45.0520 β = . 4.14) Consider a supersonic flow past a compression corner with 20 θ = . The upstream properties are 1 3.0 M = and 2 1 2116 / P lb ft = . A Pitot tube is inserted in the flow downstream of the corner.
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