UNIT - III NORMAL & OBLIQUE SHOCKS

UNIT - III NORMAL & OBLIQUE SHOCKS

1. ME 6604 – GAS DYNAMICS AND JET PROPULSION UNIT – III SHOCK WAVES Mrs. N. PREMALATHA ASSOCIATE PROFESSOR DEPARTMENT OF MECHANICAL ENGG

2. Significance of Compressible Flows Isentropic Flow a. Entropy remains constant b. All the stagnation parameters remain constant Fanno Flow a. Entropy changes (Non-isentropic process) b. Stagnation temperature remains constant Rayleigh Flow a. Entropy changes (Non-isentropic process) b. Stagnation temperature changes Flow Through Shock Waves a. Entropy increases (Non-isentropic process) b. Stagnation temperature constant Note: Isentropic flow tables are used for All the flows. How?

3. Difference between Normal and Oblique Shock Waves Normal Shock Wave Oblique Shock Wave Occurs in supersonic flow (M 1 >1) Occurs in supersonic flow (M 1 >1, No difference) Normal to the Flow direction Inclined at some other angle Flow direction does not change (Flow Deflection angle = 0) Flow direction changes (Flow Deflection angle  0) The downstream of the shock wave flow is always subsonic (M 2 <1) The downstream of the shock wave flow may be subsonic or supersonic (M 2 <1 or M 2 > 1) Shock strength is high Shock strength is low

4. Applications of Shock Waves Supersonic flow over aerodynamic bodies- Supersonic airfoils, wings, fuselage and other parts of airplanes Supersonic flow through engine components- Engine intake, Compressors, combustion chambers, Nozzle Propellers running at high speeds Current Research Design of supersonic combustor for scramjet engine Shock interactions Shock boundary layer interaction

5. Supersonic Flow Turning For normal shocks, flow is perpendicular to shock no change in flow direction How does supersonic flow change direction, i.e., make a turn either slow to subsonic ahead of turn (can then make gradual turn) = bow shock M 1 M 2 M 1 >1 M 1 >1 go through non-normal wave with sudden angle change , i.e., oblique shock (also expansions: see later ) Can have straight/curved, 2-d/3-d oblique shocks will examine straight, 2-d oblique shocks

6. Oblique Shock Waves Mach wave consider infinitely thin body M 1 >1 Oblique shock consider finite-sized wedge, half-angle,  M 1 >1  no flow turn required  infinitessimal wave  flow must undergo compression if attached shock  oblique shock at angle  similar for concave corner  M 1 >1 

7. Equations of Motion Governing equations same approach as for normal shocks use conservation equations and state equations Conservation Equations mass, energy and momentum this time 2 momentum equations - 2 velocity components for a 2-d oblique shock Assumptions steady flow (stationary shock), inviscid except inside shock, adiabatic, no work but flow work

8. Control Volume Pick control volume along shock p 1 , h 1 , T 1 ,  1 v 1   p 2 , h 2 , T 2 ,  2 v 2  - Divide velocity into two components one tangent to shock, v t one normal to shock, v n Angles from geometry v 1n v 2n v 1n v 2n A n A t p 1 p 2  1  2 v 1t v 2t v 1t v 2t v 1t v 2t v 1n = v 1 sin ; v 1t =v 1 cos  v 2n = v 2 sin ( -); v 2t =v 2 cos( -) M 1n = M 1 sin ; M 1t =M 1 cos  M 2n = M 2 sin ( -); M 2t =M 2 cos( -)

9. Conservation Equations Mass v 1n v 2n A n A t p 1 p 2  1  2 v 1t v 2t v 1t v 2t Momentum tangent normal (1) (2)

10. Conservation Equations ( con’t ) Energy v 1n v 2n A n A t p 1 p 2  1  2 v 1t v 2t v 1t v 2t (3) Eq’s . (1)-(3) are same equations used to characterize normal shocks ( with v n  v ) So oblique shock acts like normal shock in direction normal to wave v t constant, but M t1 M t2 (4) 1

11. Oblique Shock Relations To find conditions across shock, use M relations from normal shocks, but replace M 1  M 1 sin  M 2  M 2 sin( -) Mach Number from p 1 , h 1 , T 1 ,  1 v 1   p 2 , h 2 , T 2 ,  2 v 2  - v 1n v 2n v 1t v 2t (5)

12. Oblique Shock Relations ( con’t ) Static Properties p 1 , h 1 , T 1 ,  1 v 1   p 2 , h 2 , T 2 ,  2 v 2  - v 1n v 2n v 1t v 2t (6) (from pressure ratio of NS wave) (7) (from density ratio of NS wave) (8) (from temperature ratio of NS wave)

13. Oblique Shock Relations ( con’t ) Stagnation Properties T o (from energy conservation) (9)

14. Use of Shock Tables Since just replacing M 1  M 1 sin  M 2  M 2 sin( -) can also use normal shock tables use M 1 ' = M 1 sin  to look up property ratios M 2 = M 2 '/ sin( -) , with M 2 ' from normal shock tables Warning do not use p 1 /p o2 from tables only gives p o2 associated with v 2n , not v 2t p 1 , h 1 , T 1 ,  1 v 1   p 2 , h 2 , T 2 ,  2 v 2  - v 1n v 2n v 1t v 2t

15. Example #1 Given: Uniform Mach 1.5 air flow (p=50 kPa , T=345K) approaching sharp concave corner. Oblique shock produced with shock angle of 60 ° Find: T o2 p 2  (turning angle) Assume:  =1.4, steady, adiabatic, no work, inviscid except for shock,…. =60 ° M 1 =1.5 T 1 =345K p 1 =50kPa  M 2

16. Example #1 ( con’t ) Analysis: Determination of T o calculate normal component Determination of p 2 Determination of   v 1t v 1 v 1n - v 2 v 2n v 2t NOTE: Supersonic flow okay after oblique shock =60 ° M 1 =1.5 T 1 =345K p 1 =50kPa  M 2

17. Wave/Shock Angle Generally, wave angle  is not given , rather know turning angle  Find relationship between M 1 , , and    v 2 - v 2n v 1  v 1n v 1t v 2t (from relation between V1n and V2n and Vel triangle) 1 (10)

18. Oblique Solution Summary If given M 1 and turning angle,  Find  from (iteration) or use oblique shock charts Calculate M 1n =M 1 sin  Use normal shock tables or Mach relations Get M 2 from M 2 =M 2n /sin( - )  M 1  M 2

19. Example #2 Given: Uniform Mach 2.4, cool, nitrogen flow passing over 2-d wedge with 16 ° half-angle . Find: , p 2 / p 1 , T 2 /T 1 , p o2 / p o1 , M 2 Assume:  =1.4, steady, adiabatic, no work, inviscid except for shock,…. M 1 =2.4  M 2

20. Example #2 ( con’t ) Analysis: Determination of  use shock relations calculate normal component Supersonic after shock =16 ° M 1 =2.4  M 2 iterate

21. Example #2 ( con’t ) Analysis ( con’t ): use shock relations calculate normal component =16 ° M 1 =2.4  M 2 a second solution for  in addition to 39.4  Eqn generally has 2 solutions for  : Strong and Weak oblique shocks Now subsonic after shock previous solution 2.535 1.335 0.9227 1.75

22. Strong and Weak Oblique Shocks As we have seen, it is possible to get two solutions to equation (10) 2 possible values of  for given (,M 1 ) e.g., M 1 =2.4 39.4 ° M 2 =1.75 =16 ° M 1 =2.4 82.1 ° M 2 =0.575 =16 ° Examine graphical solution

23. Graphical Solution Weak shocks smaller   min = =1.4 M 2 <1 M 2 >1 Strong Weak = sin -1 (1/M 1 ) usually M 2 >1 Strong shocks  max =90 ° (normal shock) always M 2 <1 Both for  =0 no turn for normal shock or Mach wave

24. Which Solution Will Occur? Depends on upstream versus downstream pressure e.g., back pressure p 2 /p 1 large for strong shock small for weak shock Internal Flow can have p 2 much higher or close to p 1 M>1 p 1 p b,high M>1 p 1 p b,low External Flow downstream pressure usually close to upstream p (both near p atm ) M >1 p atm p atm

25. Maximum Turning Angle Given M 1 , no straight oblique shock solution for >  max (M) =1.4  max (M=3)~34 ° Given , no solution for M 1 < M 1,min Given fluid (), no solution for any M 1 beyond  max e.g., ~45.5° (=1.4)  max ()

26. Detached Shock What does flow look like when no straight oblique shock solution exists? detached shock/bow shock, sits ahead of body/turn M 1 >1 >  max bow shock can cover whole range of oblique shocks (normal to Mach wave) normal shock at centerline (flow subsonic to negotiate turn); curves away to weaker shock asymptotes to Mach wave M 2 <1 M 2 >1 M 1 >1 >  max

27. CRITICAL FLOW AND SHOCK WAVES M CR Sharp increase in c d is combined effect of shock waves and flow separation Freestream Mach number at which c d begins to increase rapidly called Drag-Divergence Mach number