Oblique Shocks -1 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 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 M1M1 M2M2 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

Oblique Shocks -2 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 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 

Oblique Shocks -3 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 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

Oblique Shocks -4 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Control Volume Pick control volume along shock p 1, h 1, T 1,  1 v1v1   p 2, h 2, T 2,  2 v2v2  -- 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 AnAn AtAt p1p1 p2p2 11 2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(  -  )

Oblique Shocks -5 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Conservation Equations Mass v 1n v 2n AnAn AtAt p1p1 p2p2 11 22 v 1t v 2t v 1t v 2t Momentum tangent normal (1) (2)

Oblique Shocks -6 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Conservation Equations (con’t) Energy v 1n v 2n AnAn AtAt p1p1 p2p2 11 22 v 1t v 2t v 1t v 2t (3) Eq’s. (1)-(3) are same equations used to characterize normal shocks (VII.1-3) with v n  v So oblique shock acts like normal shock in direction normal to wave –v t constant, but M t1  M t2 (VII.20) 1

Oblique Shocks -7 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Oblique Shock Relations To find conditions across shock, use M relations from normal shocks, e.g., (VII.5-17) but replace M 1  M 1 sin  M 2  M 2 sin(  -  ) Mach Number (VII.21) from (VI.31) p 1, h 1, T 1,  1 v1v1   p 2, h 2, T 2,  2 v2v2  -- v 1n v 2n v 1t v 2t

Oblique Shocks -8 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Oblique Shock Relations (con’t) Static Properties p 1, h 1, T 1,  1 v1v1   p 2, h 2, T 2,  2 v2v2  -- v 1n v 2n v 1t v 2t (VII.22) (from VII.12) (VII.23) (from VII.18) (VII.24) (from VII.9)

Oblique Shocks -9 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Oblique Shock Relations (con’t) Stagnation Properties T o (from energy conservation) p o (from VII.14) since function of static property ratios, don’t have to factor in p ot v. p on (VII.25) (from VII.13)

Oblique Shocks -10 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 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 v1v1   p 2, h 2, T 2,  2 v2v2  -- v 1n v 2n v 1t v 2t

Oblique Shocks -11 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 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.T o2 2.p 2 3.  (turning angle) Assume: TPG/CPG with  =1.4, steady, adiabatic, no work, inviscid except for shock,….  =60° M 1 =1.5 T 1 =345K p 1 =50kPa  M2M2

Oblique Shocks -12 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Example #1 (con’t) Analysis: –To–To –calculate normal component –p2–p2 –– (from VII.20)  v 1t v1v1 v 1n -- v2v2 v 2n v 2t NOTE: Supersonic flow okay after oblique shock  =60° M 1 =1.5 T 1 =345K p 1 =50kPa  M2M2

Oblique Shocks -13 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Wave/Shock Angle Generally, wave angle  is not a given, rather know turning angle  Find relationship between M 1, , and  (VII.26)   v2v2 -- v 2n v1v1  v 1n v 1t v 2t (from VI.43) (from geometry) 1 So to find oblique shock solution, need 2 indep. parameters, e.g., M 1 and 

Oblique Shocks -14 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Oblique Solution Summary If given M 1 and turning angle,  1.Find  from (iteration) VII.26 or use oblique shock charts (e.g., Appendix C in John) 2.Calculate M 1n =M 1 sin  3.Use normal shock tables or Mach relations, e.g., VII to get property ratios 4.Get M 2 from M 2 =M 2n /sin(  -  ) or VII.21  M1M1  M2M2

Oblique Shocks -15 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 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: N 2 is TPG/CPG with  =1.4, steady, adiabatic, no work, inviscid except for shock,…. M 1 =2.4  M2M2

Oblique Shocks -16 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Example #2 (con’t) Analysis: –– –use shock relations calculate normal component Supersonic after shock (from VII.26)  =16° M 1 =2.4  M2M2 iterate

Oblique Shocks -17 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Example #2 (con’t) Analysis (con’t): –use shock relations calculate normal component  =16° M 1 =2.4  M2M2 –a second solution for  in addition to 39.4  VII.26 generally has 2 solutions for  : Strong and Weak oblique shocks Now subsonic after shock previous solution

Oblique Shocks -18 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Strong and Weak Oblique Shocks As we have seen, it is possible to get two solutions to equation (VII.26) –2 possible values of  for given ( ,M 1 ) –e.g., M 1 = ° M 2 =1.75  =16° M 1 = ° M 2 =0.575  =16° Examine graphical solution

Oblique Shocks -19 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 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

Oblique Shocks -20 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 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 p1p1 p b,high M>1 p1p1 p b,low External Flow –downstream pressure usually close to upstream p (both near p atm ) M >1 p atm p atm

Oblique Shocks -21 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 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 (  )

Oblique Shocks -22 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 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  >  ma x