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UNIT - III NORMAL & OBLIQUE SHOCKS

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UNIT - III NORMAL & OBLIQUE SHOCKS

  1. 1. ME 6604 – GAS DYNAMICS AND JET PROPULSION UNIT – III SHOCK WAVES Mrs. N. PREMALATHA ASSOCIATE PROFESSOR DEPARTMENT OF MECHANICAL ENGG
  2. 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. 3. Difference between Normal and Oblique Shock Waves Normal Shock Wave Oblique Shock Wave Occurs in supersonic flow (M1 >1) Occurs in supersonic flow (M1 >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 (M2 <1) The downstream of the shock wave flow may be subsonic or supersonic (M2 <1 or M2 > 1) Shock strength is high Shock strength is low
  4. 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
  5. 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 M1 M2 M1>1 M1>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. 6. Oblique Shock Waves • Mach wave – consider infinitely thin body M1>1  • Oblique shock – consider finite-sized wedge, half-angle,  M1>1  – no flow turn required   1 1 M/1sin  – infinitessimal wave  – flow must undergo compression – if attached shock  oblique shock at angle  – similar for concave corner  M1>1 
  7. 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. 8. Control Volume • Pick control volume along shock p1, h1, T1, 1 v1   p2, h2, T2, 2 v2  - • Divide velocity into two components – one tangent to shock, vt – one normal to shock, vn • Angles from geometry v1n v2n v1n v2nAn At p1 p2 1 2 v1t v2t v1t v2t v1t v2t – v1n= v1sin; v1t=v1cos – v2n= v2sin(-); v2t=v2cos(-) – M1n= M1sin; M1t=M1cos – M2n= M2sin(-); M2t=M2cos(-)
  9. 9. Conservation Equations • Mass v1n v2nAn At p1 p2 1 2 v1t v2t v1t v2t   Adnv0 rel CS    2 A v 2 A vAv 2 A v 2 A vAv t t22 t t11nn22 t t22 t t11nn11   • Momentum    CS rel CS AdnvvAdnp  tangent        nn22t2nn11t1 t 21 t 21 AvvAvv 2 A pp 2 A pp  t2t1 vv  normal    nn22n2nn11n1n2n1 AvvAvvApAp  n22n11 vv  (1) 2 n22 2 n1121 vvpp  (2)
  10. 10. Conservation Equations (con’t) • Energy v1n v2nAn At p1 p2 1 2 v1t v2t v1t v2t                   2 v hAv 2 v hAv 2 2 2nn22 2 1 1nn11 2 v 2 v h 2 v 2 v h 2 t2 2 n2 2 2 t1 2 n1 1  2 v h 2 v h 2 n2 2 2 n1 1  (3) • Eq’s. (1)-(3) are same equations used to characterize normal shocks ( with vnv) • So oblique shock acts like normal shock in direction normal to wave – vt constant, but Mt1Mt2  11t 22t 1t 2t av av M M (4)2 1 1t 2t T T M M  1
  11. 11. Oblique Shock Relations • To find conditions across shock, use M relations from normal shocks, • but replace M1  M1 sin M2  M2 sin(-) • Mach Number   1sinM 1 2 1 2 sinMsinM 22 1 22 1 22 2                  1M 1 2 1 2 MM 2 1 2 1 2 2                 from p1, h1, T1, 1 v1   p2, h2, T2, 2 v2  - v1n v2n v1t v2t (5)
  12. 12. Oblique Shock Relations (con’t) • Static Properties p1, h1, T1, 1 v1   p2, h2, T2, 2 v2  - v1n v2n v1t v2t (6)1 1 sinM 1 2 p p 22 1 1 2       (from pressure ratio of NS wave) (7)     2sinM1 sinM1 v v 22 1 22 1 1 2 n2 n1       (from density ratio of NS wave)    121sinM 1sinM 1 2 sinM 2 1 1 T T 222 1 22 1 22 1 1 2                     (8) (from temperature ratio of NS wave)
  13. 13. Oblique Shock Relations (con’t) • Stagnation Properties 1o2o TT  To (from energy conservation)    12 1 211o2o ppTTpp                                 1 1 22 1 1 22 1 22 1 1o 2o 1 1 sinM 1 2 sinM 2 1 1 sinM 2 1 p p (9)
  14. 14. Use of Shock Tables • Since just replacing M1  M1sin M2  M2sin(-) – can also use normal shock tables – use M1'=M1sin to look up property ratios – M2= M2'/sin(-), with M2' from normal shock tables • Warning – do not use p1/po2 from tables – only gives po2 associated with v2n, not v2t p1, h1, T1, 1 v1   p2, h2, T2, 2 v2  - v1n v2n v1t v2t
  15. 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: 1. To2 2. p2 3.  (turning angle) • Assume: =1.4, steady, adiabatic, no work, inviscid except for shock,…. =60° M1=1.5 T1=345K p1=50kPa  M2
  16. 16. Example #1 (con’t)• Analysis: Determination of To 30.160sin5.1sinMM 1n1   191.1 805.1 786.0)( 12 12 2    TT pp MTableNS n – calculate normal component    K5005.12.01K345 M 2 1 1TTT 2 2 111o2o          Determination of p2     kPa3.90kPa50805.1pppp 1122  Determination of     2.11 191.160cos5.1 786.0 tan60 5.0 1             t2n2 MMtan     211n221t1n2 TTcosMMTTMM   v1t v1 v1n - v2 v2n v2t   104.1sinMM n22 NOTE: Supersonic flow okay after oblique shock =60° M1=1.5 T1=345K p1=50kPa  M2
  17. 17. Wave/Shock Angle • Generally, wave angle  is not given, rather know turning angle  • Find relationship between M1, , and       22cosM 1sinMtan2 tan 2 1 22 1      v2 - v2n v1  v1nv1t v2t     2sinM1 sinM1 v v 22 1 22 1 n2 n1    (from relation between V1n and V2n and Vel triangle)     tanv tanv t2 t1 1 (10)
  18. 18. Oblique Solution Summary • If given M1 and turning angle,  1. Find  from (iteration) or use oblique shock charts 2. Calculate M1n=M1sin 3. Use normal shock tables or Mach relations 4. Get M2 from M2=M2n/sin(-) M1  M2
  19. 19. Example #2 • Given: Uniform Mach 2.4, cool, nitrogen flow passing over 2-d wedge with 16° half-angle. • Find: , p2/p1 , T2/T1 , po2/po1 , M2 • Assume: =1.4, steady, adiabatic, no work, inviscid except for shock,…. M1=2.4  M2
  20. 20. Example #2 (con’t) • Analysis: Determination of  52.14.39sin4.2sinMM 1n1   9227.0;335.1;535.2;6935.0)1.( 1212122  oon ppTTppMNSTable – use shock relations calculate normal component   75.1sinMM n22  Supersonic after shock      22cos4.14.2 1sin4.2tan2 16tan 2 22    =16°M1=2.4  M2 iterate  4.39
  21. 21. Example #2 (con’t) • Analysis (con’t): 38.21.82sin4.2sinMM 1n1   5499.0;018.2;425.6;5256.0)( 1212122  oon ppTTppMNStable – use shock relations calculate normal component   575.0sinMM n22  =16°M1=2.4  M2 – a second solution for       22cos4.14.2 1sin4.2tan2 16tan 2 22    in addition to 39.4  1.82 • 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. 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 (,M1) – e.g.,       22cosM1sinMtan2tan 2 1 22 1  M1=2.4 39.4° M2=1.75 =16° M1=2.4 82.1° M2=0.575 =16° • Examine graphical solution
  23. 23. Graphical Solution • Weak shocks –smaller  –min= 0 10 20 30 40 50 60 70 80 90 0 10 20 30 40 50  (deg) (deg) 0 10 20 30 40 50 60 70 80 90 M1= 1.25 1.5 2 2.5 3 5 10 100 =1.4M2<1 M2>1 Strong Weak =sin-1(1/M1) – usually M2>1 • Strong shocks – max=90° (normal shock) – always M2<1 • Both for =0 – no turn for normal shock or Mach wave
  24. 24. Which Solution Will Occur? • Depends on upstream versus downstream pressure – e.g., back pressure – p2/p1 large for strong shock small for weak shock • Internal Flow – can have p2 much higher or close to p1 M>1 p1 pb,high M>1 p1 pb,low • External Flow – downstream pressure usually close to upstream p (both near patm) M>1 patm patm
  25. 25. Maximum Turning Angle • Given M1, no straight oblique shock solution for >max(M) 0 10 20 30 40 50 60 70 80 90 0 10 20 30 40 50  (deg) (deg) 0 10 20 30 40 50 60 70 80 90 M1= 1.25 1.5 2 2.5 3 5 10 100 =1.4 max(M=3)~34° • Given , no solution for M1<M1,min • Given fluid (), no solution for any M1 beyond max e.g., ~45.5° (=1.4) max()
  26. 26. Detached Shock • What does flow look like when no straight oblique shock solution exists? – detached shock/bow shock, sits ahead of body/turn M1>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 M2<1 M2>1 M1>1 >ma x
  27. 27. CRITICAL FLOW AND SHOCK WAVES 0.1 DivergenceDragCR MM MCR • Sharp increase in cd is combined effect of shock waves and flow separation • Freestream Mach number at which cd begins to increase rapidly called Drag- Divergence Mach number

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