Normal and Oblique Shocks, Prandtl Meyer Expansion

Aerothermodynamics of high speed ﬂows AERO 0033–1 Lecture 3: Normal and oblique shocks, Prandtl Meyer expansion

Thierry Magin, Greg Dimitriadis, and Adrien Crovato [email protected]

Aeronautics and Aerospace Department von Karman Institute for Fluid Dynamics Aerospace and Mechanical Engineering Department Faculty of Applied Sciences, University of Li`ege

Wednesday 9am – 12:15pm February – May 2021

1 / 36 Outline

Normal shock

Total and critical quantities

Oblique shock

Prandtl-Meyer expansion

Shock reﬂection and shock interaction

2 / 36 Outline

Normal shock

Total and critical quantities

Oblique shock

Prandtl-Meyer expansion

Shock reﬂection and shock interaction

2 / 36 Normal shock relations (inviscid ﬂow) Rankine-Hugoniot relations for steady normal shock

σ = 0, v = u ex , n = ex → σn(U2 − U1) = (F · n)2 − (F · n)1

ρ2u2 = ρ1u1 2 2 ρ2u2 + p2 = ρ1u1 + p1 ρ2u2H2 = ρ1u1H1

I Eqs. also valid for non calorically perfect gases I Contact discontinuity when u2 = u1 = 0 → p2 = p1 I Otherwise, the total enthalpy conservation can be expressed as

2 2 2 u cp p γ p2 u2 γ p1 u1 H2 = H1 with H = h+ , h = → + = + 2 R ρ γ − 1 ρ2 2 γ − 1 ρ1 2

I Non-linear algebraic system of 3 eqs. in 3 unknowns ρ2, u2, and p2 with closed solution expressed in function of dimensionless parameters: Mach√ number M1 = u1/a1 and speciﬁc heat ratio γ, with a1 = γRT1 3 / 36 Normal shock relations for calorically perfect gases

For M1 > 1 (derivation given further in this section)

2 ρ2 u1 (γ+1)M1 I ρ = u = 2 > 1 1 2 2+(γ−1)M1 p2 = 1 + 2γ (M2 − 1) > 1 I p1 γ+1 1 γ−1 2 2 1+ 2 M1 I M2 = 2 γ−1 < 1 γM1 − 2

h i h 2 i T2 p2 ρ2 2γ 2 2+(γ−1)M1 I T = ( p )/( ρ ) = 1 + γ+1 (M1 − 1) 2 > 1 1 1 1 (γ+1)M1

4 / 36 Entropy across normal shock The entropy variation through a normal shock is T2 p2 p2/p1 s2 − s1 = cp ln − R ln = cv ln γ ≥ 0 for M1 ≥ 1 T1 p1 (ρ2/ρ1) 2nd law of thermodynamics for calorically perfect gas: a shock wave may only happen if M1 ≥ 1 (compression shock)

I Origin of the entropy increase through a shock wave? Answer: changes across shockwaves occur through a distance of the order of the mean-free-path. Gradients are very large in the shock structure. In turns, heat ﬂuxes and viscous stresses are dissipative phenomena that generate entropy 2 I Weak shock: M1 = 1 + ε with 0 < ε 1 γ p2 = 1 + 2γ ε and ( ρ2 )−γ = (1 + ε)−γ 1 + γ−1 ε I p1 γ+1 ρ1 γ+1 p2 ρ2 −γ 2 γ(γ−1) 3 4 ( ) ∼ 1 + 2 ε + O(ε ) I p1 ρ1 3 (γ+1) s2−s1 2 γ(γ−1) 3 4 ∼ 2 ε + O(ε ) cv 3 (γ+1) Isentropic approximation valid for weak shocks I An expansion shock (M1 < 1 and M2 > 1) can be predicted

theoretically for some speciﬁc non-ideal ﬂuids 5 / 36 Outline

Normal shock

Total and critical quantities

Oblique shock

Prandtl-Meyer expansion

Shock reﬂection and shock interaction

5 / 36 Total quantities I Consider a ﬂuid element in an arbitrary ﬂow travelling at velocity v with static pressure p and temperature T

Total pressure p0 and total temperature T0 These quantities are obtained by isentropically decelerating the H = c T = c T + 1 |v|2 ﬂow to rest p 0 p 2 s = cp ln T0 − R ln p0 = cp ln T − R ln p

0 1 1

1 0 4 7 1 4 4 4 14 6 4 4 10 8 1 0 1 1 01 47

0 6 1 014 1 01 5 1 46 14 1014 10 5

Pressure ﬁeld around a Pitot probe in a plasma jet at M= 0.1 (∆p = 1 Pa)

I Pitot probes are widely used in ﬂuid dynamics (airspeed of aircraft and air and gas velocities in industrial applications) I The stagnation pressure is measured at the probe nose pPitot I Pitot probes allow to measure the stagnation pressure 6 / 36 Total quantities (cont’d)

I Total quantities are local quantities with deﬁnition still valid for viscous ﬂows d I Total enthalpy eq.: ρ dt (H) − ∂t p + ∇·q + ∇·(τ·v) = 0 I This. relation is derived using E = H − p/ρ and d p d p d ρ ( ) = (p) − (ρ) = ∂ p + v·∇p + p∇·v dt ρ dt ρ dt t

= ∂t p + ∇·(pv)

I In steady inviscid ﬂows, along a trajectory (pathline) d I The total enthalpy is conserved: dt (H) = 0 I The total temperature is constant I Through a normal shock I The total temperature is constant: T0,2 = T0,1 p The total pressure is a decreasing function: 0,2 = exp(− s2−s1 ) I p0,1 R

7 / 36 Total quantities (cont’d) I The total temperature is obtained from 1 2 H = cpT0 = cpT + 2 |v| 2 2 T0 1 v γR v γ−1 2 = 1 + 2 = 1 + 2c 2 =1+ 2 M T cpT p a

p0 p I The isentropic relations γ = γ and perfect gas law ρ0 ρ p0 = p yield the total pressure and density ρ0T0 ρT p γ 0 = 1 + γ−1 M2 γ−1 p 2 ρ 1 0 = 1 + γ−1 M2 γ−1 ρ 2 1 2 1 v 2 γ 2 I Dynamic pressure: pdyn = 2 ρv = 2 γ γRT ρRT = 2 M p I For M 1, the Bernoulli relation for incompressible ﬂows is retrieved by using Taylor’s expansion (1 + x)α = 1 + αx + O(x 2) p 0 = 1 + γ M2 + O(M4) → p ∼ p + 1 ρv 2 p 2 0 2

8 / 36 Exercise: wind-tunnel measurement You are to measure the pressure using a Pitot probe in a supersonic wind tunnel operating with air. Calculate: 1. The static pressure in the test section in [N/m2] 2. The pressure difference in [cm Hg] between the column of mercury in a U-tube manometer between the pressure measured by the Pitot probe and the wall orifice (which is used to measure the static pressure in the test section) 3. The dynamic pressure of the free stream flow

5 2 Data: pressure in the reservoir p01 = 6.0 × 10 [N/m ], free stream Mach number M1 = 3.5, barometric pressure ∆h = 75.2 [cm Hg], and 3 3 (ρg)Hg = 133.3 × 10 [N/m ] 9 / 36 Solution 1. Static pressure γ p01 = 1 + γ−1 M2 γ−1 I p1 2 1 2 → p1 = 7867 [N/m ] [Pa] 2. Manometer pressure γ−1 2 2 1+ 2 M1 I Mach number: M2 = 2 γ−1 → M2 = 0.451 γM1 − 2 p2 2γ 2 Static pressure: = 1 + (M − 1) → p2 = 111 115 [Pa] I p1 γ+1 1 γ p02 γ−1 2 γ−1 Total pressure: = 1 + M → p02 = 127 756 [Pa] I p2 2 2 I Pressure diﬀerence p02 − p1 = (ρg)Hg ∆h ⇒ ∆h = 89 cm 3. Dynamic pressure 1 γ p = ρ u2 = M2p = 67 455 [Pa] dyn1 2 1 1 2 1 1 6= p01 − p1 = 592 133 [Pa]

Rem 1: the Bernoulli relation cannot be used in compressible ﬂows Rem 2: the barometric pressure data is not used in this exercise!

10 / 36 Critical quantities I Consider a fluid element in an arbitrary flow travelling at velocity v with static pressure p and temperature T Critical conditions Let us imagine this fluid element is adiabatically decelerated (if M > 1) or accelerated (if M <√ 0) until its velocity equals the speed of sound: |v|∗ = a∗ = γRT ∗. The velocity reached at critical condition is sonic

∗ 1 ∗2 cp 1 ∗2 1 γ+1 ∗2 I By definition: H = cpT + 2 a = ( γR + 2 )a = 2 γ−1 a I Critical conditions are reached at the throat of a nozzle when the flow is chocked I In steady inviscid flows, along a trajectory (pathline) d I The total enthalpy is conserved: dt (H) = 0 I The critical temperature and speed of sound are constant I Through a normal shock I The critical temperature and speed of sound are constant ∗ ∗ ∗ ∗ T2 = T1 and a2 = a1 11 / 36 Prandtl’s relation for normal shocks ∗2 I Prandtl’s relation: a = u1u2 I Total enthalpy conservation u2 u2 γ p2 + 2 = γ p1 + 1 = 1 γ+1 a∗2 γ−1 ρ2 2 γ−1 ρ1 2 2 γ−1 Alternative form: pi = γ−1 γ+1 a∗2 − u2 , i ∈ {1, 2} I ρi 2γ γ−1 i I Introducing the previous relations in the ratio of momentum to mass shock relations 2 2 ρ2u2 +p2 ρ1u1 +p1 p2 p1 = → u2 + = u1 + ρ2u2 ρ1u1 ρ2u2 ρ1u1 yields Prandtl’s relation after some algebra

I The critical Mach number M∗ = u/a∗, can be obtained from

1 u2 γ + 1 a2 + = 1 a∗2 γ − 1 2 2 γ − 1 1 1 1 γ + 1 1 + = 1 γ − 1 M2 2 2 γ − 1 M∗2 (γ + 1)M2 M∗2 = 2 + (γ − 1)M2

12 / 36 Application of Prandtl relation and critical Mach number I The critical Mach number behaves as the local Mach number but remains ﬁnite at high speeds M∗ < 1 if M < 1 2 ∗ ∗2 (γ + 1)M M = 1 if M = 1 M = → ∗ 2 + (γ − 1)M2 M > 1 if M > 1 ∗ γ+1 M → γ−1 if M → ∞ I Alternative form of Prandtl relation for normal shocks ∗ ∗ M1 M2 = 1 I For a normal shock ∗ ∗ M1 > 1 → M1 > 1 → M2 < 1 → M2 < 1

I The normal shock relations are then easily derived as follows 2 2 2 ρ2 = u1 = u1 = u1 = M∗2 = (γ+1)M1 I ρ u u u a∗2 1 2 1 2 1 2 2+(γ−1)M1 2 2 1+ γ−1 M2 M∗2 = 1 → (γ+1)M2 = 2+(γ−1)M1 → M2 = 2 1 I 2 M∗2 2+(γ−1)M2 (γ+1)M2 2 2 γ−1 1 2 1 γM1 − 2 2 2 2 u2 I p2 − p1 = ρ1u − ρ2u = ρ1u1(u1 − u2) = ρ1u (1 − ) 1 2 1 u1 2 p2−p1 2 u2 2 (γ+1)M1 2γ 2 p = γM1 (1 − u ) = γM1 1 − 2 = γ+1 (M1 − 1) 1 1 2+(γ−1)M1 2 T2 p2 ρ2 h 2γ 2 i 2+(γ−1)M1 I T = ( p )/( ρ ) = 1 + γ+1 (M1 − 1) 2 1 1 1 (γ+1)M1 13 / 36 Outline

Normal shock

Total and critical quantities

Oblique shock

Prandtl-Meyer expansion

Shock reﬂection and shock interaction

13 / 36 Steady 2D discontinuity Rankine-Hugoniot jump relations for steady 2D discontinuity

σ = 0, v = un n + ut t → σn(U2 − U1) = (F · n)2 − (F · n)1

ρ2un2 = ρ1un1 2 2 ρ2un2 + p2 = ρ1un1 + p1 ρ2un2ut2 = ρ1un1ut1

ρ2un2H2 = ρ1un1H1

I un1 = un2 = 0: contact discontinuity, slip line ⇒ p2 = p1 I un1, un2 6= 0: oblique shock ⇒ ut2 = ut1 and H2 = H1

ρ2un2 = ρ1un1 ut2 ut1 ⇒ tan(β − θ) = tan β ρ2/ρ1

14 / 36 Oblique shock relations

1 2 2 1 γ+1 ∗2 I The total enthalpy is H = h + 2 (un + ut ) = 2 γ−1 a I The jump relations ρ2un2 = ρ1un1 2 2 ρ2un2 + p2 = ρ1un1 + p1 1 2 1 2 1 2 h2 + 2 un2 = h1 + 2 un1 = H − 2 ut are identical to the normal shock relations ⇒ For a given shock angle β and incoming Mach number M1, all ﬂow quantities behind the shock can be directly computed by the 1D normal shock relations 2 ρ2 un1 (γ+1)Mn1 p2 2γ 2 I ρ = u = 2 and p = 1 + γ+1 (Mn1 − 1) 1 n2 2+(γ−1)Mn1 1 2 T2 p2 ρ2 h 2γ 2 i 2+(γ−1)Mn1 I T = ( p )/( ρ ) = 1 + γ+1 (Mn1 − 1) 2 1 1 1 (γ+1)Mn1 γ−1 2 2 1+ 2 Mn1 I Mn2 = 2 γ−1 γMn1− 2 based on normal components Mn1 = M1 sin β and Mn2 = M2 sin(β − θ)

15 / 36 Shock polar

I The θ − β − M1 polar relation can be derived based on trigonometry

tan β tan(β − θ) = ρ2/ρ1 " # tan β − tan θ 2 + (γ − 1)M2 sin2 β = α tan β, with α = 1 2 2 1 + tan β tan θ (γ + 1)M1 sin β 1 − α tan θ = tan β 1 + α tan2 β ⇒ 2 2 M1 sin β − 1 tan θ = 2 2 cot β M1 (γ + cos 2β) + 2

16 / 36 17 / 36 Maximum deﬂection angle for oblique shocks

For any given M1, there is a maximum deﬂection angle θmax . If the physical geometry is such that θ > θmax , then no solution exists for a straight oblique shockwave. Instead the shock will be curved and detached

18 / 36 Strong shock and weak shock solutions

For any given θ < θmax , there are 2 values of β predicted

I Changes across the shock are more severe for the larger value of β (strong shock) I The weak shock is favoured and usually occur unless, for instance, the back pressure is increased by some independent mechanism

I In the weak shock solution, M2 > 1 except for a small region near θmax I If θ = 0, then β = π/2 (normal shock) or β = µ (Mach wave, evanescent shock)

19 / 36 Detached shock in front of a blunt body

The shape of the detached shock in front of a blunt body wave can be obtained by means of computational ﬂuid dynamics simulations

e Evanescent shock (zero deflection: sin β = 1/M1, Mach wave) d Weak shock, M2 > 1 c’ Weak shock, M2 = 1 c Maximum flow deflection point dividing weak (M2 > 1) and strong shock solutions b Strong shock a Normal shock 20 / 36 Prandtl’s relation for oblique shocks

∗∗2 ∗2 γ−1 2 I A normal Bernoulli constant a = a − γ+1 ut is introduced to express conservation of the normal component of the total enthalpy 1 2 1 2 1 γ + 1 ∗∗2 h2 + u = h1 + u = a 2 n2 2 n1 2 γ − 1 ∗∗2 I Prandtl’s relation for oblique shocks is a = un1un2

I A critical normal Mach number can be deﬁned as ∗ ∗∗ Mn = un/a , and Prandtl’s relation for oblique shocks is expressed as ∗ ∗ Mn1Mn2 = 1

I Notice that Mt1 6= Mt2 since a1 6= a2 I The critical tangential Mach number deﬁned as ∗ ∗∗ ∗ ∗ ∗∗ Mt = ut /a is constant: Mt1 = Mt2 since a = constant

21 / 36 Outline

Normal shock

Total and critical quantities

Oblique shock

Prandtl-Meyer expansion

Shock reﬂection and shock interaction

21 / 36 Shock wave and Mach wave

X-15 model ﬁred into ballistic wind Mach waves in supersonic nozzle tunnel at M= 3.5 [NASA] [Meyer, 1908]

22 / 36 Expansion fan

I Expansion fan formed by Mach waves I Supersonic ﬂow deﬂected downward on convex geometry (e.g. nozzle throat) I Flow properties change smoothly and continuously (6= shock wave)

I Prandtl-Meyer eq. for ﬂow angle θ Trigonometric π v + dv sin − µ cos µ convention: θ ≥ 0 = 2 = v sin π − (µ + |dθ|) cos(µ + |dθ|) counter clockwise 2 dv cos µ 1 + = v cos µ cos |dθ| − sin µ sin |dθ| 1 ∼ ∼ 1 + tan µ |dθ| 1 − tan µ |dθ| dv |dθ| ∼ tan µ |dθ| = q v 1 − 1 sin2 µ p dv dθ = − M2 − 1 v General relation also valid for non sin µ = 1 I M calorically perfect gases 23 / 36 Prandtl-Meyer function I Change of variables for calorically perfect gas 1 2 1 1 1 2 H = cpT + v = ( + )v = constant 2 (γ−1) M2 2 1 dM dH = ( 1 + 1 )2vdv − 2 v 2 = 0 (γ−1) M2 2 γ−1 M3 √ dv 1 dM − M2 − 1 dM = → dθ = v γ−1 2 M γ−1 2 M 1 + 2 M 1 + 2 M I Mach number for a Mach wave given by ν(M) = ν(M1) + θ1 − θ = ν(M1) + |θ − θ1| I The Prandtl-Meyer function is an increasing function of the q q √ γ+1 γ−1 2 2 Mach number: ν(M) = γ−1 arctan γ+1 (M − 1) − arctan M − 1

I Hypersonic limit: s π γ + 1 lim ν(M) = ν(∞) = ( − 1) M→∞ 2 γ − 1  ◦ 149 : atoms = 130◦ : linear molecules  90◦ : nonlinear molecules

24 / 36 Prandtl-Meyer expansion

I Supersonic flow over convex corner I Corner angle θ2 (fream stream θ1 =0) I Expansion wave fan centered at point A I All flow properties through expansion wave change smoothly and continuously I Exception: wall streamline changes discontinuously at point A I Flow streamlines deflected downward I Expansion fan: continuous expansion region composed of an infinite number of Mach waves and bounded by 2 Mach waves I Flow behind expansion wave: uniform and k to θ2 direction ν(M2) = ν(M1) + θ1 − θ2 = ν(M1) + |θ2 − θ1| I M2 > M1 → µ2 < µ1 I p2 < p1, ρ2 < ρ1, T2 < T1 I The expansion is isentropic, thus p02 = p01 and γ γ−1 ! p 1 + M2 γ−1 2 = 2 1 p γ−1 2 1 1 + 2 M2 25 / 36 Maximum turning angle I When the end Mach number approaches infinity, the maximum turning angle

|θmax − θ1| = ν(∞) − ν(M1)

I When the deﬂection angle exceeds the maximum angle, the ﬂow behaves as if there is almost a maximum angle and that region beyond becomes a vortex street

T0 I Notice that lim T (M) = lim = 0 M→∞ M→∞ γ−1 2 1 + 2 M I In practice, the gas becomes liquid when the temperature reaches the condensation limit 26 / 36 Outline

Normal shock

Total and critical quantities

Oblique shock

Prandtl-Meyer expansion

Shock reﬂection and shock interaction

26 / 36 Shock polar in hodograph plane (exact oblique shock theory) I The shock polar in the hodograph plane 2 21/2 (u2, v2) = (V2 cos θ, V2 sin θ), with V2 = u2 + v2 is derived from Prandtl’s relation: ∗2 γ−1 2 un1un2 = a − γ+1 ut

un1 = u1 sin β v2 v2 un2 = un1 − = u1 sin β − cos β cos β ut = u1 cos β

2 2 ∗2 γ−1 2 2 ⇒ u1 sin β − u1v2 tan β = a − γ+1 u1 cos β ∗ ∗ I Deﬁning non-dimensional velocitiesu ¯2 = u2/a ,v ¯2 = v2/a , ∗ andu ¯1 = u1/a , this relation is expressed as u¯ u¯ − 1 v¯2 = (¯u − u¯ )2 1 2 2 2 1 2 2 1 + γ+1 u¯2 − u¯1u¯2

27 / 36 I Proof Considering the relation tan β = u1−u2 , obtained based on the grey I v2 triangle, and the trigonometric relations cos2 β = 1 and 1+tan2 β 2 sin2 β = tan β , one gets 1+tan2 β (u − u )2 u2v 2 u2 1 2 − u (u − u ) = a∗2 − γ−1 1 2 1 2 2 1 1 2 γ+1 2 2 v2 + (u1 − u2) v2 + (u1 − u2) I Dividing by a∗2 2 2 2 2 2 2 γ−1 2 2 u¯1 (¯u1 − u¯2 ) − u¯1(¯u1 − u¯2)[¯v2 + (¯u1 − u¯2) ] =v ¯2 + (¯u1 − u¯2) − γ+1 u¯1 v¯2

I After some algebra 2 2 2 2 v¯2 (1 + γ+1 u¯2 − u¯1u¯2) = (¯u2 − u¯1) (¯u1u¯2 − 1)

I No deflection (θ = 0,v ¯2 = 0) I u2 = u1: evanescent shock (Mach line) I u¯2 = 1/u¯1: normal shock (Prandtl relation) I Deflection angle 0 < θ < θmax : 2 values for v2 I Strong shock solution (always supersonic flow) I Weak shock solution (either subsonic or supersonic flow) I Maximum deflection angle θ = θmax I Deflection angle θ > θmax : no solution (detached shock) 28 / 36 I For M1 = 2 I θ = 0◦ Point A : evanescent shock, Point F: normal shock ◦ I 0 < θ = 10 < θmax , Point B: weak shock, Point E: strong shock I θ = θmax , Point D: maximum deflection angle I Point C: separation between supersonic and subsonic regimes (the critical Mach number M∗ behaves as the local Mach number M)

29 / 36 Shock polar in pressure-deflection plane I Pressure-deflection diagram: locus of all possible static pressure behind an oblique shock wave as a function of the deflection angle

I It is a parametric curve depending on the free stream Mach number M : p2 = 1 + 2γ (M2 sin2 β − 1), where 1 p1 γ+1 1 β = β(M1, θ) I Points B and B’: weak shock solutions depending on the sign of θ I Points E and E’: strong shock solutions I Points G and G’: expansion shock (not physical) 30 / 36 Shock reﬂection from a solid boundary I Consider an oblique shock wave deviating the ﬂow from region 1 to 2 through an angle −∆θ incident on a solid wall at point I

I The streamline is deflected from region 2 to 3 through a reflected shock to leave the flow parallel to the wall ◦ I M1 = 2 and ∆θ = −10 ⇒ M2 = 1.641 and p2/p1 = 1.707 ◦ I M2 = 1.641 and ∆θ = +10 ⇒ M3 = 1.285 and p3/p1 = 2.805

I In the pressure-deﬂection plane

I M2 < M1: the reﬂected shock is weaker than the oblique shock I p3 − p2 > p2 − p1 I The shock is not specularly reﬂected

31 / 36 Approximate solution based on the characteristic theory

I On s+ characteristic between regions 1 and 2

◦ ◦ ◦ ν(M2) = ν(M1) + ∆θ = 26.379 − 10 = 16.379 ⇒ M2 = 1.651

I Isentropic relation

γ γ−1 ! p 1 + M2 γ−1 2 = 2 1 = 1.705 p γ−1 2 1 1 + 2 M2

I On s− characteristic between regions 2 and 3

◦ ◦ ◦ ν(M3) = ν(M2) − ∆θ = 16.379 − 10 = 6.379 ⇒ M3 = 1.308

I Isentropic relation

γ γ−1 ! p 1 + M2 γ−1 3 = 2 1 = 2.795 p γ−1 2 1 1 + 2 M3

32 / 36 Singular shock reflection from a solid boundary I When the deflection ∆θ is larger than the maximum deflection angle θmax for the M2 value, no regular shock

reflection is allowed ◦ I M1 = 2 and ∆θ = −16 ⇒ M2 = 1.403 and p2/p1 = 2.308 ◦ I M2 = 1.403 and ∆θ = +16 ⇒ no oblique shock solution I A normal shock is formed at the wall to allow the streamlines to continue parallel to the wall (Mach stem) I Away from the wall, this normal shock transits into a curved shock which intersects the incident shock I Singular reflexion: the reflected shock is curved in reality I Slip line Σ between regions 3 and 4 results from different entropy jump I Intersection between two polars: ◦ p3 = p4 (< p5) ⇒ ∆θ < 16 I Subsonic region: no analytical result I Triple point T: incident & reflected shocks + Mach stem (+slip line) 33 / 36 Shock interaction I Consider two oblique shocks of the same intensity deviating the flow from region 1 to 2 through an angle |∆θ| and interacting at point I

I The streamlines are deﬂected from region 2 to 3 through refracted shocks to leave the ﬂow parallel to the symmetry plane

◦ I M1 = 1.95 and ∆θ = −10 ⇒ M2 = 1.594 and p2/p1 = 1.694 ◦ I M2 = 1.594 and ∆θ = +10 ⇒ M3 = 1.233 and p3/p1 = 2.783

I In the pressure-deﬂection plane I M2 < M1: the refracted shock is weaker than the oblique shock I p3 − p2 > p2 − p1 I No slip line due to the symmetry of the problem

34 / 36 Approximate solution based on the characteristic theory

I On s+ characteristic between regions 1 and 2

◦ ◦ ◦ ν(M2) = ν(M1)+∆θ = 24.992 −10 = 14.992 ⇒ M2 = 1.604

I On s− characteristic between regions 2 and 3

◦ ◦ ◦ ν(M3) = ν(M2)−∆θ = 14.992 −10 = 4.992 ⇒ M3 = 1.256

35 / 36 Singular shock interaction (∼ type II interference, Edney, 1968) I When the deflection ∆θ is larger than the maximum deflection angle θmax for the M2 value, no regular shock refraction is allowed ◦ I M1 = 1.95 and ∆θ = −14 ⇒ M2 = 1.440 and p2/p1 = 2.069 ◦ I M2 = 1.440 and ∆θ = +14 ⇒ no oblique shock solution I A normal shock is formed between the two incidents shocks to allow the streamlines to continue parallel to the symmetry plane (Mach disk for revolution flow) I Away from the wall, this normal shock transits into a curved shock which intersects the incident shock

I Two triple points T1 and T2 I Two slip lines Σ1 and Σ2 between regions 3 and 4 result from diﬀerent entropy jump I Intersection between two polars: ◦ p3 = p4 ⇒ ∆θ < 14 36 / 36