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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: Machp 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 + " with 0 < " 1 γ p2 = 1 + 2γ " and ( ρ2 )−γ = (1 + ")−γ 1 + γ−1 " I p1 γ+1 ρ1 γ+1 p2 ρ2 −γ 2 γ(γ−1) 3 4 ( ) ∼ 1 + 2 " + O(" ) I p1 ρ1 3 (γ+1) s2−s1 2 γ(γ−1) 3 4 ∼ 2 " + O(" ) 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 jvj2 flow to rest p 0 p 2 s = cp ln T0 − R ln p0 = cp ln T − R ln p 1 1 0 0 1 1 1 0 4 7 1 4 4 4 14 6 4 4 10 8 1 0 1 1 01 47 7 4 1 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 jvj 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 <p 0) until its velocity equals the speed of sound: jvj∗ = 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 2 f1; 2g 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 ! 1 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 .
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