Advanced Computational Fluid Dynamics AA215A Lecture 4

Transcription

1 Advaned Computational Fluid Dynamis AA5A Leture 4 Antony Jameson Winter Quarter,, Stanford, CA

2 Abstrat Leture 4 overs analysis of the equations of gas dynamis

3 Contents Analysis of the equations of gas dynamis. Coordinate Transformations Analysis of the equations of gas dynamis: the Jaobian matries Two dimensional flow One dimensional flow Transformation to alternative sets of variables: primitive form Symmetri form Riemann invariants Symmetri hyperboli form Entropy variables

4 Chapter Analysis of the equations of gas dynamis

5 Leture 4 Analysis of the equations of gas dynamis. Coordinate Transformations In order to alulate solutions for flows in omplex geometri domains, it is often useful to introdue body fitted oordinates through global, or, as in the ase of isoparametri elements, loal transformations. With the body now oiniding with a oordinate surfae, it is muh easier to enfore the boundary onditions aurately. Suppose that the mapping to omputational oordinates (ξ, ξ, ξ 3 ) is defined by the transformation matries K ij = ξ j, K ij = ξ i x j, J = det(k). (.) The Navier-Stokes equations (?? -?? ) beome Here the transformed fluxes are t (Jw) + F i (w) =. (.) ξ i F i = S ij f j, (.3) S = JK. (.4) The elements of S are the ofators of K, in a finite volume disretization they are just the fae areas of the omputational ells projeted in the x, x, x 3 diretions. Using the permutation tensor ɛ ijk we an express the elements of S as S ij = ɛ jpqɛ irs x p ξ r x q ξ s. (.5) Then S ij = ( ξ i ɛ x p x q jpqɛ irs + x p ) x q ξ r ξ i ξ s ξ r ξ s ξ i =. (.6) Defining saled ontravariant veloity omponents as U i = S ij u j, (.7) 3

6 CHAPTER. ANALYSIS OF THE EQUATIONS OF GAS DYNAMICS 4 the flux formulas may be exped as F i = U i U i u + S i p U i u + S i p U i u 3 + S i3 p U i H. (.8) If we hoose a oordinate system so that the boundary is at ξ l =, the wall boundary ondition for invisid flow is now U l =. (.9). Analysis of the equations of gas dynamis: the Jaobian matries The Euler equations for the three-dimensional flow of an invisid gas an be written in integral form, using the summation onvention, as d wdv + f dt i n i ds = Σ Σ is the domain, dσ its boundary, n the normal to the boundary, dv ds are the volume area elements. Let x i, u i,, p, E H denote the Cartesian oordinates veloity, density, pressure, energy enthalpy. In differential form dσ t + f (w) = (.) i the state flux vetors are Also, w = u u u 3 E, f i = u i u u u 3 H + p δ i δ i δ i3 p = (γ )(E u ), H = E + p = γ + u u is the speed is the speed of sound u = u i, = γp Let m i e denote the momentum omponents total energy, Then w f an be expressed as m i = u i, e = E = p γ + m i

7 CHAPTER. ANALYSIS OF THE EQUATIONS OF GAS DYNAMICS 5 w = m m m 3 e, f i = u i m m m 3 e + p δ i δ i δ i3 u i (.) In a finite volume sheme the flux needs to be alulated aross the interfae between eah pair of ells. Denoting the fae normal area by n i S, the flux is F S f = n i f i This an be expressed in terms of the onservative variables w as u n is the normal veloity f = u n m m m 3 e + p n n n 3 u n (.) Also u n = n i u i = n im i p = (γ )(e m i ) In smooth regions of the flow the equations an also be written in quasilinear form as A i are the Jaobian matries t + A i = A i = f i, The omposite Jaobian matrix at a fae with normal vetor n is A = A i n i = f All the entries in f i f are homogenous of degree in the onservative variables w. It follows that f i f satisfy the identities f i = A i w, f = Aw This is the onsequene of the fat that if a quantity q an be expressed in terms of the omponents of a vetor w as q = j w α j j, j α j = α, then

8 CHAPTER. ANALYSIS OF THE EQUATIONS OF GAS DYNAMICS 6 i q i w i = i = i α i α i q j w α j j (.3) In order to evaluate A note that un Aordingly = αq are row vetors u n = [ u n, n, n, n 3, ] [ ] u = (γ ), u, u, u 3, f = (u nw)+ n n n 3 u n [ ]+ p [ un ] (.4) Then the Jaobian matrix an be assembled as the sum of a diagonal matrix two outer produts of rank, A = u n I + [ un ] + [ u n ] u n u 3 n 3 H u n = u n I + [ u n, n, n, n 3, ] + (γ ) [ u u n, u, u, u 3, ] u n u 3 n 3 H u n Note that every entry in the Jaobian matrix an be expressed in terms of the veloity omponents u i the speed of sound sine H = γ + u It may also be diretly verified that f = Aw beause u n is homogeneous of degree with the onsequene that un is orthogonal to w. while p is homogenous of degree so that u n w = w = p

9 CHAPTER. ANALYSIS OF THE EQUATIONS OF GAS DYNAMICS 7 Thus Aw = u n w + n n n 3 u n [ ] w = f Also f = A + A w = A beause A A is homogeneous of degree, with the onsequene that w is in the null spae of. The speial struture of the Jaobian matrix enables the diret identifiation of its eigenvalues eigenvetors. Any vetor in the 3 dimensional subspae orthogonal to the vetors un is an eigenvetor orresponding to the eigenvalue u n, whih is thus a triple eigenvalue. It is easy to verify that the vetors r = u u u 3 u r = are orthogonal to both un n 3 u u n 3 u 3 n r = n 3 n u 3 n u n 3 r 3 = n n u n u n. However, r, r r 3 are not independent sine 3 n k r k = k= Three independent eigenvetors an be obtained as v = n r + r, v = n r + r, v 3 = n 3 r + r 3 is the speed of sound. In order to verify this note that the middle three elements of r k, k =,, 3 are equal to i k n, i k is the unit vetor in the k th oordinate diretion. Also the last element of v k is equal to i k ( n k u ). If the vetors v k are not independent they must satisfy the relation of the form v = 3 α k v k = k= for some non zero vetor α. The first element of v is 3 α k n k = α n k= For the next three elements of v to be zero 3 k= ( i α k k ) n = α n =

10 CHAPTER. ANALYSIS OF THE EQUATIONS OF GAS DYNAMICS 8 whih is only possible if α is parallel to n, so that α n. In order to identify the remaining eigenvetors denote the olumn vetors in A as u a = u u 3 a n = n n 3 H u n Then Now onsider a vetor of the form Then u n a =, p u n a = a =, r = a + αa a = if whih is the ase if Thus the vetors Ar = u n (a + αa ) + αa + a (.5) = λ(a + αa ) = λr u n + α = λ, αu n + = λ α = v 4 = a + a, v 5 = a a are the eigenvetors orresponding to the eigenvalues u n + u n. Written in full u + n v 4 = u + n u 3 + n 3 v u n 5 = u n u 3 n 3 H + u n H u n.3 Two dimensional flow The equations of two dimensional flow have the simpler form t + f (w) + f (w) = x x the state flux vetors are p w = m m, f = u w + p e u, f = u w + p u

11 CHAPTER. ANALYSIS OF THE EQUATIONS OF GAS DYNAMICS 9 Now the flux vetor normal to an edge with normal omponents n n is The orresponding Jaobian matrix is The eigenvalues of A are with orresponding eigenvetors v = u u v = u f = n f + n f A = f = n f + n f = u n I + u u n u + H n n u n u n, u n, u n +, u n n n (u n u n ).4 One dimensional flow v 3 = u + n u + n H + u n The equations of one dimensional flow are further simplified to the state vetor is the flux vetor is f = Now the Jaobian matrix is A = f = ui + u u + H The eigenvalues of A are t + x f (w) = w = u E u u + p uh with the orresponding eigenvetors v = u u v = u = m e = uw + = ui + u, u +, u u + H + u u H u up v 4 = [ u,, ] + (γ ) v 3 = u H u u n u n H u n u [ u, u, ]

12 CHAPTER. ANALYSIS OF THE EQUATIONS OF GAS DYNAMICS.5 Transformation to alternative sets of variables: primitive form The quasi-linear equations of gas dynamis an be simplified in various ways by transformations to alternative sets of variables. Consider the equations of three dimensional flow t + A i = A i = f i Under a transformation to new set of variables w Multiplying by w = M, In the ase of the primitive variables, we find that M = M w t + A i M = w M w = w t + w Ãi = Ã i = M A i M, Ai = M = w = u u u 3 p MÃi M u u u 3 u u u u 3 γ u u u 3 (γ ) u (γ )u (γ )u (γ )u γ u p u A = u u u

13 CHAPTER. ANALYSIS OF THE EQUATIONS OF GAS DYNAMICS A = A 3 = u p u u u u u 3 p u 3 u 3 u 3 u 3.6 Symmetri form Consider the equations of one dimensional flow in primitive variables t + u x + u x = u t + u u x + x = u + t x + u x = Then subtrating the first equation multiplied by from the third equation, we find that + u( t t x x ) = This is equivalent to a statement that the entropy is onstant sine If the entropy is onstant, then S = log( p ) = log p γ log γ ds = dp p γ d = p (dp d) dp = d With this substitution the first equation beomes, t + u x + u x = now the first two equations an be resaled as t + u x + u x = u t + x + u u x =

14 CHAPTER. ANALYSIS OF THE EQUATIONS OF GAS DYNAMICS Thus if we write the equations in terms of the differential variables we obtain the symmetri form d w = w t Ā = dp du dp d + Ā w x = u u u These transformations an be generalized to the equations of three dimensional flow. A onvenient saling is to set dp d w = du du du (.6) 3 d dp This eliminates the density from the transformation matries M = dw d w M = d w dw. The equations now take the form w t + w Āi = (.7) the transformed Jaobian matries Ā i = M A i M (.8) are simultaneously symmetrized as Ā = Ā = Ā 3 = u u u u u u u u u u u 3 u 3 u 3 u 3 u 3 (.9) (.) (.)

15 CHAPTER. ANALYSIS OF THE EQUATIONS OF GAS DYNAMICS 3 while The ombined Jaobian matrix an now be deomposed as M = w = u u u u u 3 u 3 H u u u 3 u γ u γu γu γu 3 γ M = w = u u u 3 γ(u H) γu γu γu 3 γ (.) (.3) γ = γ (.4) A = n i A i (.5) A = n i MĀ i M (.6) Corresponding to the fat that Ā is symmetri one an find a set of orthogonal eigenvetors, whih may be normalized to unit length. Then one an express Ā = V Λ V the diagonal matrix Λ ontains the eigenvalues u n, u, u n, u n + u n as its elements. The matrix V ontaining the orresponding eigenvetors as its olumns is V = n n n 3 n n n n 3 n n 3 n 3 n n n n 3 V = V T. The Jaobian matrix an now be expressed as (.7) A = MΛM (.8) This deomposition is often useful. Sine it follows that in isentropi flow M = M V, M = V T M = dp d d = γ γp d d = γ d d = γ dp (.9)

16 CHAPTER. ANALYSIS OF THE EQUATIONS OF GAS DYNAMICS 4 Thus dp = γ d so the equations an also be expressed in the same form, equations (.7), (.8), (.9), (.) (.) for the variables γ u w = u u 3 S with the transformation matries M = dw d w = M = d w dw =.7 Riemann invariants p pu u u pu u 3 pu 3 H u u u 3 p u γ u γu γu γu 3 γ u u u 3 γ p (u H) γu p γu p γu 3 p γ = γ In the ase of one dimensional isentropi flow these equations redue to γ t + u t + u γ x + u x = γ x + u u x = S t + u S x = Now the first two equations an be added subtrated to yield γ p R + t R t + (u + ) R+ x = + (u ) R x =

17 CHAPTER. ANALYSIS OF THE EQUATIONS OF GAS DYNAMICS 5 R + R are the Riemann invariants R + = u + γ, R = u γ whih remain onstant as they are transported at the wave speeds u + u. The Riemann invariants prove to be useful in the formulation of far field boundary onditions designed to minimize wave refletion..8 Symmetri hyperboli form Equation (.7) an be written in terms of the onservative variables as Here w = M. Now multiplying by form w t + Āi M T the w = equation is redued to the symmetri hyperboli Q t + Âi = (.3) Q is symmetri positive definite the matries Âi are symmetri Q = (MM T ), Âi = M T Ā i M (.3) This form ould alternatively be derived by multiplying the onservative form (.) of the equations by Q. Then Q f i = M T M f = M T M A i = M T M MĀi = M T A i M M A symmetri hyperboli form an atually be obtained for any hoie of the dependent variables. For example equation (.7) ould be written in terms of the primitive variables as N w t + ĀiN w = N = w w = Then multiplying by N T we obtain the symmetri hyperboli form N T N w t + N T Ā i N w =

18 CHAPTER. ANALYSIS OF THE EQUATIONS OF GAS DYNAMICS 6.9 Entropy variables A partiular hoie of variables whih symmetrizes the equations an be derived from funtions of the entropy S = log( p ) = log p γ log (.3) γ The last equation of (.7) is equivalent to the statement that S t + u S i = whih an be ombined with the mass onservation equation multiplied by S to yield the entropy onservation law S t + S (u i ) = (S) t + (u is) = (.33) This is a speial ase of a generalized entropy funtion defined as follows. Given a system of onservation laws t + f (w) = (.34) i Suppose that we an find a salar funtion U(w) suh that U f i = G i (.35) U(w) is a onvex funtion of w. Then U(w) is an entropy funtion with an entropy flux G i (w) sine multiplying equation (.34) by U we obtain using (.35) this equivalent to U t + U f i = (.36) U(w) t + G i = whih is in turn equivalent to the generalized entropy onservation law U(w) t + G i = (.37) Now introdue dependent variables v T = U Then the equations are symmetrized. Define the salar funtions (.38) Q(v) = v T w U(w) (.39)

19 CHAPTER. ANALYSIS OF THE EQUATIONS OF GAS DYNAMICS 7 Then R i (v) = v T f i G i (w) (.4) Q v = w T + v T v U v = w T R i v = f T i + v T f i v G i v = f T i Hene v is symmetri. j v k = Q v j v k = k v j f i v is symmetri f ij v k = R i v j v k = f ik v i orrespondingly f i v = A i v is symmetri. Thus the onservation law (.34) is onverted to the symmetri form v v t + A i v v = (.4) Multiplying the onservation law (.34) by v from the left produes the alternative symmetri form v t + v A i = (.4) sine v A i = v (A i v ) v Moreover, multiplying (.34) by v T reovers equation (.36) the orresponding entropy onservation law (.37). Conversely, if U(w) satisfies the onservation law (.37), then but These an both hold for all only if U t = G i U t = U f i U f i = G i so u(w) is an entropy funtion if it is a onvex funtion of w.

20 CHAPTER. ANALYSIS OF THE EQUATIONS OF GAS DYNAMICS 8 The property of onvexity enables U(w) to be treated as a generalized energy, sine u beomes unbounded if w beomes unbounded. It an be shown that S is a onvex funtion of w. Aordingly it follows from the entropy equation (.33) that we an take U(w) = S γ = γ γ log as a generalized entropy funtion the saling fator expressions. Using the formula we find that γ = (γ )[u, u, u, u 3, ] v T = u [ γ S = u γ p, u p, u p, u 3 p, ] p u = u i u i The reverse transformation is w T = p [ v 5, v, v 3, v 4, ( (v + v 3 + v 4)v ] 5 log p (.43) γ is introdued to simplify the resulting Now the salar funtions Q(v) R i (v) defined by equations (.39) (.4) redue to Aordingly (.44) (.45) Q(v) =, R i (v) = m i = u i (.46) w T = v = v f T i = m i v = m i = [ ] v m i = [ δ i δ i δ i3 ] Thus the first four rows of v onsist of w T, f T, f T f T 3, orrespondingly sine v is symmetri its first four olumns are w, f, f f 3. The only remaining element is 5 v 5. A diret alulation reveals that 5 v 5 = E + (E + u )p

21 CHAPTER. ANALYSIS OF THE EQUATIONS OF GAS DYNAMICS 9 Thus v = u u u 3 E u u + p u u u 3 u u E + u p u u u u + p u 3u u E + u p u 3 u u 3 u u 3 u 3 + p u 3E + u 3 p E u E + u p u E + u p u 3 E + u 3 p E + (E + u = wwt + p u u u 3 u u u 3 E + u )p