Oblique Shocks Less Irreversible Thermodynamic Devices The pessimist

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

Hypersonic Flyers : Comprehensive Mechanism of Intake Air Control

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

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

Oblique Shock is a Barbed Wire Fence

Though Experiment : Tangential Component of Momentum (- rxuxwx A + ry uy wy A ) = 0 • But from continuity wx = wy rxux = ry uy Tangential velocity is Constant across an oblique Shock wave

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

Energy Equation Steady Adiabatic Flow Write Velocity in terms of components 2 Vx = ux 2 + wx 2 & Vy 2 = uy 2 + wy 2 Tangential component of velocity is not responsible for energy conversion • thus … 2 2 ux uy = hy + hx + 2 2

Collected Oblique Shock Equations • Continuity r xu x = r y u y • Momentum wx = w y 2 2 px + rxux = py + ry uy • Energy ux 2 uy 2 = c p Ty + c p Tx + 2 2 wx b b-q ux uy b-q wy q

An oblique sock is a normal Shock to Normal Velocity Component Vx & Mx & w y V & My y y • Defining: Mn x Vx & Mx y Mt • Then by similarity we can write the solution Mnx=Mxsin( Mtx=Mxcos( My & uy & Mn wx & ux x Mt Vy & Mny = æ g - 1) ( 2ö çè 1 + 2 Mnx ÷ø æ g - 1)ö ( 2 çè g Mnx - 2 ÷ø

The Normal Component of Tamed Devil All the scalar quantities change only due to normal velocity g + 1)Mnx 2 ry ( = rx (2 + (g - 1)Mnx 2 ) • Similarity Solution 2 g py 2 = 1+ Mnx - 1 (g + 1) px ( ) 2 g Ty é 2 = ê 1 + Mnx - 1 Tx ë (g + 1) ( Letting change in ( ù é 2 + (g - 1)Mnx 2 úê 2 g + ( ) n 1 M ê x ûë ) Mnx = M x sin ( ) )ùú úû

M ny = Then …. . æ (g - 1) 2ö (M x sin ) ÷ çè 1 + ø 2 æ (g - 1)ö 2 çè g (M x sin ) ÷ 2 ø g + 1)(Mx sin ) ry ( = rx 2 + (g - 1)(Mx sin )2 2 ( ( ) ) 2 py = + 2 g ( 1 Mx sin ) - 1 (g + 1) px ( ) 2 é ù ê 2 + (g - 1)(Mx sin ) 2 Ty = é + 2 g ( Mx sin ) - 1 ú ê ê 1 2 Tx ë (g + 1) g + û ë ( 1)(Mx sin ) ( )ùú ú û • Change in Properties across Oblique Shock wave ~ f(Mx, )

Total Mach Number Downstream of Oblique Shock • Consider the geometry of down stream flow Mt y b-q Mn y My 3 M Mny My = sin ( - q )

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

Oblique Shock Wave Angle Chart Mx=5. 0 Mx=3. 0 max curve Mx=1. 5 Mx=2. 0 Mx=2. 5 Mx=4. 0 Mx

Limiting Cases of Oblique Shock Wave qmax q

qmax Maximum Turning Angle

Highest Angle Objects

Performance of An Adiabatic Oblique Shock 2 g py 2 = 1+ Mnx - 1 (g + 1) px ( ) 2 g Ty é 2 = ê 1 + Mnx - 1 Tx ë (g + 1) But ( ( ù é 2 + (g - 1)Mnx 2 úê 2 g + ( ) n 1 M x û êë ) )ùú úû 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: My 3 q My 2 Mx My 1 q q