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Mechanism of Supplying Controlled Air to Hypersonic Flyers

Prof. Subbarao explains equations for oblique shocks and their effects on supersonic and hypersonic flyers, detailing the continuity, momentum, and energy equations. Learn about the relationship between normal and tangential components and how to calculate key parameters.

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Mechanism of Supplying Controlled Air to Hypersonic Flyers

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  1. Oblique Shocks : Less Irreversible Thermodynamic Devices “The pessimist complains about the Shocks; the optimist expects it to change; the realist adjusts the flyers.” P M V Subbarao Professor Mechanical Engineering Department Means to Supply Controlled Air to Supersonic Flyers !!!

  2. Hypersonic Flyers :Comprehensive Mechanism of Intake Air Control

  3. Continuity Equation for Oblique Shock Wave wy Vy wx Vx uy ux x y • For Steady Flow thru an Oblique Shock

  4. • For Steady Flow w/o Body Forces Momentum Equation for Oblique Shock Wave wy Vy wx Vx uy ux x y

  5. Oblique Shock is a Barbed Wire Fence

  6. ( ) - r + r = 0 u w A u w A x x x y y y r = r u u = w w x x y y x y Though Experiment : Tangential Component of Momentum • But from continuity Tangential velocity is Constant across an oblique Shock wave

  7. Normal Component of Momentum Equation Normal component is always subjected to normal shock!!!

  8. = + & = + 2 2 2 2 2 2 V u w V u w x x x y y y 2 2 u u + = + x y h h x y 2 2 Energy Equation Steady Adiabatic Flow Write Velocity in terms of components Tangential component of velocity is not responsible for energy conversion • thus …

  9. r = r u u x x y y b w x = w w x y b-q u + r = + r 2 2 p u p u x x x x y y y u w y y 2 2 u u b-q + = + x y c T c T q x y p p 2 2 Collected Oblique Shock Equations • Continuity • Momentum • Energy

  10. Vx & Mx Vy & My wx & Mtx ux & Mnx uy & Mny wy & Mty ( ) Vy & My æ ö g - 1 + 2 1 M n ç ÷ Vx & Mx x è ø 2 = M n ( ) y æ ö g - 1 g - 2 M n ç ÷ x è ø 2 An oblique sock is a normal Shock to Normal Velocity Component • Then by similarity we can write the solution • Defining: Mnx=Mxsin(b Mtx=Mxcos(b

  11. ( ) g + r 2 1 M n = x y ( ) ( ) r + g - 2 2 1 M n x x g 2 p ( ) ( ) = b = + - s i n 2 M M y 1 1 M n n ( ) x g + x x 1 p x ( ) ( ) é ù + g - 2 é ù 2 1 M g 2 n T ( ) x = + - 2 ê ú y 1 1 M ê ú n ( ) ( ) x g + g + 2 1 1 T M ê ú n ë û ë û x x The Normal Component of Tamed Devil All the scalar quantities change only due to change in normal velocity • Similarity Solution Letting

  12. ( ) æ ö g - 1 ( ) 2 + b 1 s i n M ç ÷ x è ø 2 = M n ( ) y æ ö g - 1 ( ) 2 g b - s i n M ç ÷ x è ø 2 ( ) ( ) 2 g + b r 1 s i n M = x y ( ) ( ) ( ) r 2 + g - b 2 1 s i n M x x ( ) g 2 p ( ) 2 = + b - y 1 s i n 1 M ( ) x g + 1 p x ( ) ( ) ( ) é ù 2 + g - b 2 1 s i n M ( ) é ù g 2 T ( ) ê ú x 2 = + b - y 1 s i n 1 M ê ú ( ) ( ) ( ) ê ú x g + 2 1 T g + b 1 s i n M ë û x ë û x Then ….. • Change in Properties across Oblique Shock wave ~ f(Mx, )

  13. Mn y Mt y M M b-q 3 y M n = y M ( ) y b - q s i n Total Mach Number Downstream of Oblique Shock • Consider the geometry of down stream flow

  14. Determination of Oblique Shock Wave Angle • Properties across Oblique Shock wave ~ f(Mx, ) • q is the geometric angle that “forces” the flow thru OS. • How do we relateqto b 

  15. Mx=5.0 max curve Mx=4.0 Oblique Shock Wave Angle Chart Mx=3.0 Mx=1.5 Mx=2.5 Mx=2.0 Mx

  16. Limiting Cases of Oblique Shock Wave qmax q

  17. Maximum Turning Angle qmax

  18. Highest Angle Objects

  19. g 2 p ( ) = + - 2 y 1 1 M n ( ) x g + 1 p x ( ) ( ) é ù + g - 2 é ù 2 1 M g 2 n T ( ) x = + - 2 ê ú y 1 1 M ê ú n ( ) ( ) x g + g + 2 1 1 T M ê ú n ë û ë û x x Performance of An Adiabatic Oblique Shock But Pressure Recovery Factor or Total Pressure Ratio for the oblique shock :

  20. Performance of An Adiabatic Oblique Shock Across a Shock Therefore Pressure Recovery Factor or Total Pressure Ratio for the oblique shock :

  21. Special Designs of Center Bodies If there are multiple shocks: My3 q My2 My1 Mx q q

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