Aerothermodynamics of high speed flows AERO 0033 1 Lecture 6: D potential flow, method of characteristics Thierry Magin, Greg Dimitriadis, and Johan Boutet Thierry.Magin@vki.ac.be Aeronautics and Aerospace Department von Karman Institute for Fluid Dynamics Aerospace and Mechanical Engineering Department Faculty of Applied Sciences, University of Liège Room B5 +1/443 Wednesday 9am 1:00pm February May 017 Magin (AERO 0033 1) Aerothermodynamics 016-017 1 / 6
Outline 1 Governing equations Method of characteristics 3 Exercises: exact theory of oblique shocks / expansion waves Magin (AERO 0033 1) Aerothermodynamics 016-017 / 6
Governing equations Outline 1 Governing equations Method of characteristics 3 Exercises: exact theory of oblique shocks / expansion waves Magin (AERO 0033 1) Aerothermodynamics 016-017 / 6
Governing equations Entropy eq. (Navier-Stokes) First law of thermodynamics for open system: T ds =de + pd( 1 ) Mass and thermal energy (Navier-Stokes) eqs. ) Entropy eq. D Dt (s) = T D (e) = pr u r q : ru Dt D ( ) Dt = r u D Dt (e) p T D Dt ( )= 1 T r q 1 T : ru Transport fluxes: = S and q = rt Conservative form: @ t ( s)+r ( us)+r ( q T )= The entropy production rate is nonnegative = T S:S + T rt 0 In smooth inviscid flows, the entropy is constant along a trajectory (pathline) D (s) =0 Dt Magin (AERO 0033 1) Aerothermodynamics 016-017 3 / 6
Governing equations Total enthalpy eq. Total energy E = e + 1 u D (E)+r (pu)+r q + r ( u) =0 Dt Total enthalpy H = h + 1 u = E + p/ p D D Dt ( p )= D Dt (p) Dt ( ) =@ tp + u rp + pr u = @ t p + r (pu) ) D Dt (H) @ tp + r q + r ( u) =0 In steady inviscid flows, the total enthalpy is constant along a trajectory (pathline) D (H) =0 Dt Magin (AERO 0033 1) Aerothermodynamics 016-017 4 / 6
Governing equations Crocco s eq. Momentum eq.: @ t u + u ru = 1 rp 1 r Lagrange identity: u ru =(r u) u + 1 r(u u) Vorticity vector:! = r u First law of thermodynamics: dh = T ds + 1 dp ) 1 rp = rh T ) @ t u +! u + 1 r(u u) =T rs rh 1 r Crocco s eq @ t u +! u = T rs rh 1 r rs In steady inviscid flows, with constant H (rh = 0), the entropy is constant in irrotational flows (! = 0)! u = T rs Conversely, if the flow is isentropic rs = 0, it must be irrotational unless! k u everywhere (Beltrami flow) In smooth inviscid and steady flows, the entropy vanishes on a pathline: u rs = 0. This property does not imply that rs =0 Magin (AERO 0033 1) Aerothermodynamics 016-017 5 / 6
Governing equations Rotational flows Exemples of rotational flow: boundary layer detached shock! = r u 6= 0)rs 6= 0 Magin (AERO 0033 1) Aerothermodynamics 016-017 6 / 6
Governing equations Irrotational flows Steady flow Uniform upstream conditions Isentropic flow Inviscid flow (no dissipation) No discontinuities ) Potential (irrotational) flow! = r u =0 u = r Exemples of irrotational flow: attached shock (except shock region), nozzle, expansion Magin (AERO 0033 1) Aerothermodynamics 016-017 7 / 6
Governing equations Alternative form of continuity eq. D Continuity eq.: Dt ( )+ r u =0 Let us consider a smooth inviscid and steady flow The material derivative reads D Dt ( ) =u r The entropy is conserved on a path line: u rs =0 By definition of the speed of sound: u r = 1 a u rp Multiplying by u the momentum eq.: ) D Dt ( ) = a u u : ru u rp = u ( u ru) = u u : ru Alternative form of the continuity eq. valid for both rotational and irrotational flows a r u u u : ru =0 Magin (AERO 0033 1) Aerothermodynamics 016-017 8 / 6
Method of characteristics Outline 1 Governing equations Method of characteristics 3 Exercises: exact theory of oblique shocks / expansion waves Magin (AERO 0033 1) Aerothermodynamics 016-017 8 / 6
Method of characteristics D potential flow: method of characteristics Non linear potential eq. a r u u u : ru =0 Irrotational flow! = r u =0 ) @ x u 1 @ x1 u =0 ) System of partial di erential eqs. a u1 u 1 u u1 @ 0 1 x1 u with Q = u1 u + u1 u a u 1 0 @ x1 Q + A @ x Q =0 1 A = a u1 u1 @ x =0 u u 1 u a u (a u1 ) 0 Magin (AERO 0033 1) Aerothermodynamics 016-017 9 / 6
Method of characteristics Eigen vectors and eigen values Left eigen vector l and eigen value : l A = l Characteristic equation: l (A I) = 0 ) A I = 0 (a u 1) Discriminant: = a (u1 + u a ) M > 1! > 0 : hyperbolic eq. M =1! = 0 : parabolic eq. M < 1! < 0:ellipticeq. For M > 1(" = ±) +u 1 u + a u =0 Real eigen values: " = u1u/a +" p M 1 u1 /a 1 Eigen vectors: (l 11 l 1 ) A = + (l 11 l 1 )and(l 1 l ) A = (l 1 l ) l 1 /l 11 =( + a 11 )/a 1 and l /l 1 =( a 11 )/a 1 After some algebra, one gets l 1 /l 11 = and l /l 1 = + Magin (AERO 0033 1) Aerothermodynamics 016-017 10 / 6
Method of characteristics Physical interpretation of the characteristic lines Change of variables u1 /a = M cos u /a = M sin " = M sin cos + " p M 1 M cos 1 = M tan + " p M 1(1 + tan ) (" p M 1) tan = ("p M 1tan +1)(" p M 1+tan ) (" p M 1 tan )(" p M 1+tan ) = = tan + "/ p M 1 1 " tan / p M 1 : 1/ cos =1+tan tan +tan("µ) 1 tan tan("µ) : sinµ =1/M, tan µ =1/p M 1 ) " =tan( + "µ) with positive in the counterclockwise direction Limits: lim M!1 µ =0andlim M!1 + µ = / Magin (AERO 0033 1) Aerothermodynamics 016-017 11 / 6
Method of characteristics tan " = " =tan( + "µ) The streamline at one point makes an angle with the x 1 axis Two characteristics are passing at this point: one at the angle µ above the streamline, and the other at the angle µ below the streamline The characteristic lines are Mach lines No discontinuity in the velocity or any other fluid property along the characteristics, Mach lines are patching lines for continuous flows Magin (AERO 0033 1) Aerothermodynamics 016-017 1 / 6
Method of characteristics Characteristic eqs. for M > 1 + 0 l11 l Using LA = L with = and L = 1 0 l 1 l The system of partial di erential eqs. reads L@ x1 Q + LA @ x Q =0) L@ x1 Q + L @ x Q =0 After explicit calculation l11 (@ x1 + + @ x )u 1 + l 1 (@ x1 + + @ x )u = 0 l 1 (@ x1 + @ x )u 1 + l (@ x1 + @ x )u = 0 The terms between ( ) can be interpreted as directional derivatives in the x 1, x plane. Using the characteristic directions s " = (cos ", sin " ) ( d d l11 u ds + 1 + l 1 u ds + = 0 d d l 1 u ds 1 + l u ds = 0 one obtains a system of ordinary di erential eqs. (hodograph plane) along the directions s + and s ) ( l11 + l 1 du du 1 = 0 on s + l 1 + l du du 1 = 0 on s in the u 1, u plane Magin (AERO 0033 1) Aerothermodynamics 016-017 13 / 6
Method of characteristics Solution method Cauchy problem: u 1 and u are given on an arc compute the flow downstream of that arc and one wishes to Discretized solution The characteristic slope direction + and can be computed at each point of arc, in particular + A =tan + A and B =tan B The position of point P is defined by the intersection of the characteristic lines approximately numerically by their tangent at A and B The velocity components at P can be computed by integrating (numerically) the characteristic equations ( u1 P u 1 A u 1 P u s + + P u A A s + = 0 on s + u 1 B u s + + P u B B s = 0 on s Magin (AERO 0033 1) Aerothermodynamics 016-017 14 / 6
Method of characteristics Zone of influence and zone of silence Consider left- and right-running characteristics through point P Zone of influence for P: areabetweenthetwodownstream characteristics. This region is influenced by any disturbances or information at point P Zone of silence for P: all points outside the zone of influence will not be a ected by disturbances or information at P Zone of dependence for P: areabetweenthetwoupstream characteristics. Properties at point P depend on any disturbances or information in the flow within this upstream region In steady supersonic flow, disturbances do not propagate upstream Magin (AERO 0033 1) Aerothermodynamics 016-017 15 / 6
Method of characteristics Riemann invariants The characteristic eqs. in the hodograph plane ( 1+ du du 1 = 0 on s + 1+ + du du 1 = 0 on s have only one solution through the point of the plane These solutions are called Riemann invariants F1 (u 1, u ) = constant on s + F (u 1, u ) = constant on s After change of variables (u 1, u )! (M, ), one obtains (M) = constant on s+ (M)+ = constant on s with the the Prandtl-Meyer function (M) introduced in lecture Magin (AERO 0033 1) Aerothermodynamics 016-017 16 / 6
Method of characteristics Proof u1 = u cos ) du Change of variables 1 =cos du u sin d u = u sin ) du =sin du + u cos d du 1 + " du = 0 (cos + " sin )du + u( sin + " cos )d = 0 (cos( "µ)cos +sin( "µ)sin )du +u( cos( "µ)sin +sin( "µ)cos )d = 0 cot µ 1 du "d = 0 u 8 >< H = c pt + 1 u 1 =( ( 1)M + 1 )u Change of variable: 1 dh = ( ( 1)M >: + 1 )udu dm u 1 M 3 = 0 p M 1 1 g(m)dm "d = 0 : with g(m) = M 1+ 1 M Z M (M) " = constant : with (M) = g(m 0 )dm 0 0 Magin (AERO 0033 1) Aerothermodynamics 016-017 17 / 6
Method of characteristics Prandtl-Meyer expansion Let us consider an incoming flow following a curved (convex) surface It is possible to compute exactly the flow quantities at each point P of the wall (the angle is negative in the clockwise direction) On s emanating from the uniform incoming flow region (M P )+ P = (M A ) For any point Q on s + emanating from point P (MQ ) Q = (M P ) P (M Q )+ Q = (M A ) = (M P )+ P! (M Q )= (M P )and Q = P :theslopeofs + is identical at all point (straight line) The angle + µ decreases: the characteristics diverge and form a fan Magin (AERO 0033 1) Aerothermodynamics 016-017 18 / 6
Method of characteristics Formation of a shock wave Let us consider an incoming flow following a curved (concave) surface, the characteristics converge The characteristics from the same family might therefore intersect Such an intersection indicates the breakdown of the method of characteristics, because there would be too much information to calculate the flow quantities at the point of intersection In reality, the intersection of characteristics in not possible, it rather indicates the appearance of a shock wave As opposed to characteristics, shocks are patching lines for discontinuous flows Magin (AERO 0033 1) Aerothermodynamics 016-017 19 / 6
Method of characteristics In practice, finite-volume representation In the finite-volume representation of the method of characteristics, the solution, supposed piecewise constant over elementary cells, is obtained as follows: Divide the Cauchy arc into a number of segments on which the flow conditions are assumed to be constant Approximate by straight lines the characteristics emanating from the segment edges Based on the characteristics, define a set of elementary cells on which the flow quantities will be determined Compute the slopes of the characteristics On s : (M 3 )+ 3 = (M 1 )+ 1 On s + : (M 3 ) 3 = (M ) ) ( (M 3 ) = (M 1)+ (M ) + 1 3 = (M 1) (M ) + 1+ ) ( 13 = 1+ 3 + µ 1+µ 3 3 = + 3 µ +µ 3 Magin (AERO 0033 1) Aerothermodynamics 016-017 0 / 6
Method of characteristics When characteristics hit physical boundaries Wall: flow angle imposed Free jet: pressure and thus Mach number imposed (provided that isentropic assumption is valid) Example: wall On s + : (M 3 ) 3 = (M ) 3 = wall ) ) (M 3 )= (M )+ wall 3 = + 3 µ +µ 3 Magin (AERO 0033 1) Aerothermodynamics 016-017 1 / 6
Exercises: exact theory of oblique shocks / expansion waves Outline 1 Governing equations Method of characteristics 3 Exercises: exact theory of oblique shocks / expansion waves Magin (AERO 0033 1) Aerothermodynamics 016-017 1 / 6
Exercises: exact theory of oblique shocks / expansion waves Exercise: forces on D airfoils Consider a symmetrical diamond-shaped airfoil flying at an angle of attack to the free stream of 8 when the upstream Mach number is. The atmospheric pressure of the free stream is equal to 101,35 Pa. The ratio of the thickness t to the chord c of the airfoil is equal to 0.1. Calculate the lift and drag forces exerted on the airfoil assuming that the chord is equal to 1m (R = 87 J/ (kg K), =1.4, T 1 =188K). Magin (AERO 0033 1) Aerothermodynamics 016-017 / 6
Exercises: exact theory of oblique shocks / expansion waves Solution The airfoil half-angle is determined from =arctan(t/c) =5.7106 The pressure coe cient for region i =1,, 3, 4isdefinedas C pi = p i p 1 = 1 1u 1 M1 ( p i 1) p 1 The forces exerted on the airfoil in the local reference frame are T = (p 1 + p 3 p p 4 )l sin = M 1 p1 N = (p 1 + p p 3 p 4 )l cos = M 1 p1 t (C p1 + C p3 C p C p4 ) c (C p1 + C p C p3 C p4 ) Magin (AERO 0033 1) Aerothermodynamics 016-017 3 / 6
Exercises: exact theory of oblique shocks / expansion waves The lift and drag forces exerted on the airfoil in the flow reference frame are L = T sin + N cos D = T cos + N sin Region 1: oblique shock theory Deflection angle =8 + =13.7106 Shock-wave angle: =43.6598 Normal Mach: M n1 =1.3807 Static pressure: p 1 /p 1 =.0575 Pressure coe cient: C p1 =0.377 Static temperature: T 1 T 1 = Normal Mach number: Mn1 = 1+ 1 h 1+ +1 (M n1 1) i apple +( 1)M n1 ( +1)M n1 M n1 Mn1 1 =0.7479 q Tangentiel Mach number: M t1 = M T1 t1 T1 =1.98 q Mach number: M 1 = Mn1 + M t1 =1.498 Region : Prandtl-Meyer expansion Mach number: (M )= (M 1 )+ =3.734 ) M =1.8889 1 p Static pressure: 1+ M = 1 1 p 1 1 ) p = p p 1 =1.1436 1+ M p 1 p 1 p 1 Pressure coe cient: C p =0.051 =1.4 Magin (AERO 0033 1) Aerothermodynamics 016-017 4 / 6
Exercises: exact theory of oblique shocks / expansion waves Region 3: Prandtl-Meyer expansion Mach number: (M 3 )= (M 1)+ =8.6684 ) M 3 =.0844 1 p Static pressure: 3 1+ M = 1 1 p 1 1 =0.8766 1+ M 3 Pressure coe cient: C p3 = 0.0441 Region 4: Prandtl-Meyer expansion Forces Mach number: (M 4 )= (M 3 )+ =40.0896 ) M 4 =.5417 1 p Static pressure: 4 1+ M = 1 1 p 1 1 =0.49 1+ M 4 Pressure coe cient: C p4 = 0.039 C T = 1 t c (C p1 + C p3 C p C p4 )=0.05(0.377 0.0441 0.051 + 0.039) = 0.049 C N = 1 (C p1 + C p C p3 C p4 )=0.5(0.377 + 0.051 + 0.0441 + 0.039) = 0.338 T = N = M1 p1 C T =6891N/m M1 p1 C N =95893N/m Magin (AERO 0033 1) Aerothermodynamics 016-017 5 / 6
Exercises: exact theory of oblique shocks / expansion waves L = T sin + N cos = 94 001 N/m D = T cos + N sin = 0 170 N/m For a supersonic inviscid flow over an infinite wing, the drag per unit span is finite. This wave drag is inherently related to the loss of total pressure and increase of entropy across the oblique shock waves created by the airfoil. Magin (AERO 0033 1) Aerothermodynamics 016-017 6 / 6