Supersonic Flow Turning

Transcription

1 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 go through non-normal wae with sudden angle change, i.e., oblique > shock (also expansions: see later) Can hae straight/cured, -d/3-d oblique shocks will examine straight, -d oblique shocks Oblique Shocks -

2 Oblique Shock Waes Recall ach wae consider infinitely thin body > no flow turn required infinitessimal wae μ sin ( / ) μ Oblique shock consider finite-sized wedge, half-angle, δ > flow must undergo compression to turn if attached shock oblique shock at angle > similar for concae corner Oblique Shocks - > μ δ δ

3 Equations of otion Goerning equations same approach as for normal shocks use conseration equations and state equations Conseration Equations mass, energy and momentum this time momentum equations - elocity components for a -d oblique shock Assumptions steady flow (stationary shock), iniscid except inside shock, adiabatic, no work but flow work Oblique Shocks -3

4 Control Volume p, h, Pick control olume along shock p, h, Diide elocity into two components one tangent to shock, t one normal to shock, n Angles from geometry n sin; t cos n sin(-δ); cos(-δ) n sin; t cos n sin(-δ); t cos(-δ) Oblique Shocks -4 t n -δ T, ρ δ n A n ρ A t ρ t t T, ρ n p p n

5 ass ρ n A n + ρ Conseration Equations 0 ρ rel t ρ A CS t n A r ( n) da + ρ n t + ρ A t t A ρ n ρn () t + ρ t A omentum p n da ρ( rel n) r r t da n t p p A n ρ A t ρ t CS CS At At + p p + p t ρn An + t ρnan tangent ( p ) ( ) ( ) ( ) t p A t normal p A ( ρ A ) + ( ρ ) n n n n n n nan n p Oblique Shocks -5 p ρ n + ρ n ()

6 Oblique Shocks -6 Conseration Equations (con t) Energy ρ n A n n h + t ρ n n h h h n h n + A + (3) n h t + n p p A n ρ A t ρ Eq s. ()-(3) are same equations used to characterize normal shocks (VII.-3) with n So oblique shock acts like normal shock in direction normal to wae t t a t T t constant, but t t t t a t t t T n (VII.0)

7 Oblique Shock Relations To find conditions across t shock, use relations from normal shocks, e.g., (VII.5-7) p, h, but replace sin n sin(-δ) n ach Number (after shock) γ from (VII.) + γ γ γ sin ( δ) sin + sin γ γ n -δ T, ρ δ p, h, T, ρ n (VII.) Oblique Shocks -7

8 Oblique Shock Relations (con t) Static Properties (from VII.) p γ sin p γ + (from VII.8) T T n n ρ ρ + (from VII.9) γ γ + ( γ + ) sin ( γ ) sin + γ sin sin t n p, h, -δ T, ρ (VII.) δ γ γ (VII.3) sin ( γ + ) ( γ ) (VII.4) p, h, T, ρ n Oblique Shocks -8

9 Oblique Shock Relations (con t) Stagnation Properties T o (from energy conseration) o o p o (from VII.4) Oblique Shocks -9 T T (from VII.3) p p o o p o p o γ + sin γ + sin γ γ ( T T ) ( p p ) γ γ γ γ + sin since function of static property ratios, don t hae to factor in p ot. p on γ γ + γ (VII.5)

10 Wae/Shock Angle Equations aboe are functions of, (shock angle) and δ (turning angle) Is there a relationship between them? (from VII.3) n n ( γ + ) sin ( γ ) sin + tan δ ( )( tan sin ) (from geometry) t ( γ + cos) + t tan tan ( δ ) t (VII.6) tan δ n Alternate Eq n. -δ δ n ( tan )( sin ) ( sin ) γ + So to find oblique shock solution, need indep. parameters, e.g., and δ Oblique Shocks -0

11 Use of Shock Tables Since just replacing t n sin sin(-δ) p, h, -δ T, ρ can also use normal δ shock tables use ' sin to look up property ratios '/sin(-δ), with ' from normal shock tables Warning do not use p /p o from tables only gies p o associated with n, not n p, h, T, ρ Oblique Shocks -

12 Oblique Solution Summary If gien and shock angle,. Find δ from VII.6 or use oblique shock charts (Appendix C in John). Calculate n sin 3. Use normal shock tables or ach relations, e.g., VII.-5 to get property ratios 4. Get from n /sin(-δ) or VII. If gien and turning angle, δ. Find from (iteration) VII.6 or use oblique shock charts (e.g., Appendix C in John). Steps -4 aboe δ Oblique Shocks -

13 Example # Known Shock Angle Gien: Uniform ach.5 air flow (p50 kpa, T345K) approaching sharp concae corner. Oblique shock produced with shock angle of 60 Find:. T o.5 T 345K p 50kPa 60 δ. p 3. δ (turning angle) Assume: TPG/CPG with γ.4, steady, adiabatic, no work, iniscid except for shock,. Oblique Shocks -3

14 Analysis: T o T o Oblique Shocks -4 To T + 345K Example # (con t) γ ( ( ) ) K p calculate normal component n sin.5sin p p p kPa 90.3 ( ) ( ) kpa o p δ NOTE: (from VII.6) o o tan 60.5 sin 60 tan δ.4 o.5.5 sin 60 n sin( δ).04 > ( )( ) ( ) t n -δ.5 T 345K p 60 50kPa δ (B.) p T o δ p.805 T.9 n Supersonic flow okay after oblique shock n

15 Example # Known Turn Angle Gien: Uniform ach.4, cool, nitrogen flow passing oer -d wedge with 6 half-angle..4 δ6 Find:, p /p, T /T, p o /p o, Assume: N is TPG/CPG with γ.4, steady, adiabatic, no work, iniscid except for shock,. Oblique Shocks -5

16 Example # (con t) Analysis: (from VII.6) o tan6.4 ( )( tan.4 sin ) (.4 + cos) +.4 δ6 iterate 39.4 use shock relations calculate normal component n sin.4sin o or from John App. C: oblique shock charts o ( B.) n ; p p.535; T T.335; po po 0.97 ( δ). 75 n sin Supersonic after shock Oblique Shocks -6

17 Oblique Shocks -7 Example # (con t) Analysis (con t): can find second solution for o tan6.4 in addition to use shock relations calculate normal component ( )( tan.4 sin ) (.4 + cos) + n sin.4sin 8. o.38 ( B.) n 0.556; p p 6.45; T T.08; po po ( δ) n sin o.4 δ6 preious solution Now subsonic after shock VII.6 generally has solutions for : Strong and Weak oblique shocks

18 Oblique Shocks -8 Alternate Approach Possible to find direct relationship * for as function of and δ δ γ χ πα λ tan 3 3 cos 4 cos tan (VII.7) δ Remoes iteration requirement just longer equations * e.g., Anderson, odern Comp. Flow, 3 rd ed., p. 4 ( ) δ γ γ λ tan ( ) δ γ γ γ λ χ tan 4 9 weak shock solution strong shock solution 0 α