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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 !!!.
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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 !!!
Hypersonic Flyers :Comprehensive Mechanism of Intake Air Control
Continuity Equation for Oblique Shock Wave wy Vy wx Vx uy ux x y • For Steady Flow thru an Oblique Shock
• For Steady Flow w/o Body Forces Momentum Equation for Oblique Shock Wave wy Vy wx Vx uy ux x y
( ) - 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
Normal Component of Momentum Equation Normal component is always subjected to normal shock!!!
= + & = + 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 …
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
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
( ) 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
( ) 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, )
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
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 relateqto b
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
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 :
Performance of An Adiabatic Oblique Shock Across a Shock Therefore Pressure Recovery Factor or Total Pressure Ratio for the oblique shock :
Special Designs of Center Bodies If there are multiple shocks: My3 q My2 My1 Mx q q