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MAE 362 Solution HW5

MAE 362 Solution HW5 - MAE 362 Solution HW 5 1[25pts Recall...

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Unformatted text preview: MAE 362 Solution HW # 5 1. [25pts] Recall that we developed normal shock relations for the limiting case of 1 M P (sec 3.6). Do the same for the oblique shock relations and determine relations for 2 1 P P , 2 1 T T , θ , 2 M . (For completeness I’ll provide 2 1 ρ ρ ) Our normal shock relation for pressure is, ( 29 2 2 2 1 1 2 1 sin 1 1 P M P γ β γ = +- + . If we take the limit this becomes, 2 2 2 1 1 2 sin 1 P M P γ β γ = + . Our normal shock relation for density is, ( 29 ( 29 2 2 1 2 2 2 1 1 1 sin 1 sin 2 M M γ β ρ ρ γ β + =- + . If we take the limit this becomes, 2 1 1 1 ρ γ ρ γ + =- . Finally, our normal shock relation for temperature is seen below. ( 29 ( 29 ( 29 2 2 1 2 2 2 2 1 1 2 2 1 1 2 1 1 sin 2 2 1 sin 1 1 1 sin M T P M T P M γ β ρ γ β ρ γ γ β- + = = +- + + If we take the limit we can state that, ( 29 ( 29 2 2 2 2 2 1 1 2 1 2 1 2 1 sin sin 1 1 1 T M M T γ γ γ γ β β γ γ γ-- = = + + + To analyze our angle of inclination and beta relationship, we can make certain statements that allow us to go a step further for a more simplified yet still accurate solution. The general methodology is as follows, ( 29 2 2 1 2 1 sin 1 tan 2cot cos2 2 M M β θ β γ β- = + + => ( 29 ( 29 cos sin sin 2 tan 2 cos2 cos2 β β β θ γ β γ β = = + + However, if we make the statement that the angle of inclination is small, and as a result...
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MAE 362 Solution HW5 - MAE 362 Solution HW 5 1[25pts Recall...

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