The Navier-Stokes Equation

Transcription

1 Revised October 9, 007 8:48 Copyright 000 David A. Randall David A. Randall Department of Atmospheric Science Colorado State University, Fort Collins, Colorado Analysis of the relative motion near a point Suppose that the velocity of the fluid at position r 0 " x! y! z# and time t is V" x! y! z! t#, and that the simultaneous velocity at a neighboring position r 0 + r is V +!V. We can write!u &!v ' "!w# & (u (u (u ' (x (y (z x (v (v (v (5.) (x (z (z & y ' " z# (w (w (w " (x (y (z # In vector form, this can be written as, (5.) where D is called the deformation tensor because it describes how the fluid element is being deformed by the non-uniform motion field. Using Cartesian tensor notation, we can alternatively express (5.) as ( v!v i x i, where. (5.) (x j i!! j We can decompose ( v i ( x j into two parts which are symmetric and antisymmetric in the suffices i and j. They are and!v D ) r (v * ij -- i (v & j ', (5.4) " (x j (x # i Selected Nuggets 9

2 94 (v + ij -- i (v & j ' " (x j (x # i, (5.5) respectively. The deformation D can now be expressed by D * +,, (5.6) where we define two parts of D : & e xx e xy e xz ' & 0 + xy + xz ' * - e yx e yy e yz, and, - +. (5.7) yx 0 + yz " e zx e zy e zz # " + zx + zy 0 # We show below that * is related to the divergence, and, is related to the vorticity. Correspondingly,!V can be divided into two parts, i.e.!v!v " s#!v " a# " s# " a# +, where!v i e ij X j, and!v i + ij X j. (5.8) Here the superscripts s and a denote symmetric and antisymmetric, respectively. We refer to * as the rate of strain tensor, or simply as the strain tensor. The diagonal elements of * ( u (v (w, i.e. e xx -----! e, are called the normal strains, (x yy ----! e (y zz (z and represent the rate of volume expansion, as illustrated in Fig. 5. a. The sum of the normal strains, i.e., the trace of the strain tensor, is the divergence of V, i.e. e xx + e yy + e zz e ii. ) V. (5.9) The off-diagonal elements of *, namely e xy, e yz, and e zx, are called the shearing strains, and express the rate of shearing deformation, as illustrated in Fig. 5. b. They can be written as e xy e yx -- (u (v " & (y (x# ', e yz e zy -- (v (w " & (z (y# ', e zx e xz -- (w (u " & (x (z# '. (5.0)

3 5. Analysis of the relative motion near a point 95 y a d v D y (u (v (x (y 0 Dx du x y b du D y (u (v (y (x 0 Dx d v x Figure 5.: a) The normal strains, representing the rate of volume expansion. b) The shearing strains, which represent the rate of shearing deformation. The elements of the anti-symmetric tensor,,, are + zy + yz -- &(w (v '! + " (y (z# xz + zx -- (u (w & '! + " (z (x# yx + xy (v (u " & (x (y# '. (5.) Introducing the notation is the vorticity vector / x + yz! / y + zx! / z + zx, we see that "/ x! / y! / z # /. V. (5.)

4 96 In tensor notation, (5.) can be written as ( v / i * k ijk , (x j (5.) where * ijk when i, j, k are in even permutation, and * ijk when i, j, k are in odd ( v permutation. For example, / z / *, which gives (x * ( v (v (u (x (x (y twice the angular velocity about the z-axis, as illustrated in Fig. 5.. Thus - d u 0 Dy ( v (u (x (y dv Dx Figure 5.: The component of the vorticity in the z direction. &!u " a# '!v " a# "!w " a# # & 0 / z / y ' & ' x -- /, (5.4) z 0 / x y " / x 0 # " z # / y or!v " a#. (5.5) -- " / r # 5. Invariants of the strain tensor Consider the scalar product

5 5. Invariants of the strain tensor 97 r ) * ) r r ) "* ) r# e xx x + e yy y + e zz z + " e xy xy + e yz yz + e zx zx# f" x! y! z#. (5.6) By putting f" x! y! z# constant, we obtain a surface of second order, the strain quadratic, also called the ellipsoid of strain. We refer the strain quadratic to its principal axes the form " X! X! X #. Then (5.6) takes e X + e X + e X F" X! X! X # constant. (5.7) Here e, e, and e are the principal values of the strain tensor, and are called the principal normal strains. Using (5.6), we see that!u " s# -- (f e (x xx x + e xy y + e xz z!v " s# -- (f ---- e (y yx x + e yy y + e yz z 4.!w " s# ( f e ( z zx x + e zy y + e zz z (5.8) In terms of the principal axes " X! X! X #,!v " s # -- (F " s# s e (X X!!v e X!!v " # e X. (5.9) " s# " s# " s#!v!v!v "!! # are the projections of!v s upon the principal axes. Therefore, the symmetric part of!v is composed of three expansions (or contractions) in the mutually orthogonal directions of the principal axes of the strain quadratic. We next consider the sum of the diagonal elements of e, i.e. e xx + e yy + e zz e ii. ) V. (5.0) This quantity is an invariant under the rotation of the axes of reference. Consider the " #

6 98 transformation by which the surface f" x! y! z# the constant of (5.6) is transformed into F" X! X! X # the constant of (5.7). We can write the transformation between the two Cartesian coordinates in the schematic, tabular form shown in Table 5. x y z a b g X X a b g X a b g Table 5.: The transformation between two Cartesian coordinate systems. Here ai, bi, gi, are the direction cosines between the axes. Here ai, bi, gi, are the direction cosines between the axes. We find that 6 i 6 i 0 i i i 5! 0, i j + 0 i 0 j + 7 i 7 j! ij 4 (5.) where! ij is Kronecker s delta "! ij when i j!! ij 0 when i8 j#. Substituting the expression into (5.7), we obtain X i i x + 0 i y + 7 i z where i!! (5.) 6 i i 5! 6 i 0 i 0. i i j + 0 i 0 j + 7 i 7 j! ij 4 (5.) This expression must be identical with (5.6), so that e xx e + e + e! e xy 0 e + 0 e + 0 e, 5 e yy 0 e + 0 e + 0 e! e yz 0 7 e e e, 4. e zz 7 e + 7 e + 7 e! e zy 7 e + 7 e + 7 e, (5.4)

7 5. Stress 99 It follows that 9 e xx + e yy + e zz 6 " + i #e i e i i + e + e i (5.5) i.e., the sum e xx + e yy + e zz. ) V is independent of the choice of reference. Of course, this quantity is called the divergence, and we knew already that it is independent of the choice of reference. Another invariant of the strain tensor is e xx e yy + e yy e zz + e zz e xx e xy e yz e zx constant. (5.6) The proof is left as an exercise. 5. Stress In a moving viscous fluid, forces act not only normal to a surface but also tangential to it. In Fig. 5., : x, : y and : z denote respectively the forces per unit area acting upon the surfaces normal to the positive x, y, and z directions, respectively. In z (: : z z z (z : x : x 0 (: : y y y (y y : y : y x (: : x x x (x : z : z Figure 5.: Sketch illustrating the normal and tangential forces acting on an element of fluid.

8 00 general, each force is a vector with components, so that : x : xx i + : yx j + : zx k, 5 : y : xy i + : yy j + : zy k,. : z : xz i + : yz j + : zz k. 4 (5.7) Symbolically we can write an element of the stresses as : ij : ij " i!! ; j!! #. Here is the i-component of the force per unit area exerted across a plane surface element normal to the j-direction. The stress is thus a tensor with nine elements, and can be represented by the matrix T & : xx : xy : xz ' : yx : yy : yz " : zx : zy : zz # (5.8) For an arbitrarily chosen plane surface, whose normal vector is unit area is. See Fig The equilibrium condition is : n : y : y : n 9; : x 9; x : y 9; y : z 9; z 0, n, the force acting per (5.9) where 9; is the area of the oblique surface, 9; x, 9; y, 9; z, are the areas of the faces lying in the planes x 0! y 0 and z 0. Denoting the direction cosines of n by, 9; 0 and 7, we have x 9; , y 9; , z 7. Therefore (5.9) can be written as 9; 9; 9; : n : x + 0: y + 7 : z, (5.0) or & : xn ' & : xx : xy : xz ' & ' : yn :, (5.) yx : yy : yz 0 " : zn # " : zx : zy : zz # " 7 # or : n T ) n. (5.)

9 5. Stress 0 z : n n"! 0! 7 # C B y : x : y 0 : z A x Figure 5.4: Sketch illustrating the normal force acting on a plane surface. Equation (5.) shows that matrix T transforms a vector n into another vector : n, so that T is a tensor, called the stress tensor. See Fig The three diagonal elements, : xx, : yy, and : zz, are normal stresses. The six off-diagonal elements are tangential stresses. An important property of T is its symmetry, i.e. : ij : ji. This symmetry can be deduced from the condition of moment equilibrium. Consider for example a moment about the z-axis due to : yx and : xy : ": yx 9y9z# --9x " : xy 9x9z# --9y -- ": yx : xy #9x9y9z. (5.) The shear stresses acting on the negative x- and y-surfaces contribute also, so that the total moment about the z-axis is ": yx : xy #9x9y9z. (5.4) Euler s equation (for a rotating rigid body) states that

10 0 z :< yx : yx :< xy : xy 0 : xy y 9z : yx 9x 9y x Figure 5.5: Sketch illustrating the components of the stress tensor. d I z z ": dt yx : xy #9x9y9z, (5.5) " 9x# + " 9y# > where I z x9y9z &" 9x# + " 9y# ' is the moment of inertia of the fluid element >9x9y9z, and z is the angular velocity about the positive z-axis. Then (5.5) reduces to > (" 9x# + " 9y# ) d z : dt yx : xy. (5.6) d In order that z remains finite, we need ": when. It follows dt yx : xy #? 0 9x! 9y? 0 that : yx : xy. Similar considerations establish that the stress tensor is symmetric, i.e., : xy : yx! : yz : zy! : zx : xz. (5.7)

11 5.4 Invariants of the stress tensor Invariants of the stress tensor As in the case of the strain tensor, we have a scalar product r ) T ) r : X + : X + : X constant, (5.8) where :, :, and : are the principal stresses acting on the surface elements normal to the principal axes (X, X, X ) of the ellipsoid of stress. It can be shown that r ) T ) r is an invariant under the rotation of the axes of reference. The sum of the normal stresses, i.e., the trace of the stress tensor, : xx + : yy + : zz : ii : + : + : p, (5.9) is also an invariant. For a moving fluid, (5.9) is the definition of pressure reduces to the static pressure when the fluid is at rest. p, which 5.5 The resultant forces due to the spatial variation of the stresses is The resultant force acting in the x-direction on a volume element of fluid 9x9y9z (: : xx xx x " & (x # ' 9y9z :xx 9y9z : ( : + & y xy ' 9x9z " xy (y # (: : xy 9y9z + : xz xz z " & (z # ' 9x9y :xz 9x9y (: xx (: xy (: xz " & (x (y (z # ' 9x9y9z. (5.40) Therefore the three components of the resultant force (vector) per unit volume are X Y Z (: xx (: xy (: xz (x (y (z (: yx (: yy (: yz (x (y (z (: zx (: zy (: zz (x (y (z (: xj (x j (: yj (x j (: zj (x j 5 (5.4) Equation (5.4) can be compactly written as

12 04 DivT (: x (x (: y (: z (y (z (: i xj (x j ( : yj j (y k ( : zj. j (zj (5.4) DivT is a tensor divergence and, therefore, is a vector. 5.6 Physical laws connecting the stress and strain tensors The relation between the stress tensor and the strain tensor may be established by the following assumptions. i) : ij p! ij + F ij, (5.4) where F ij is the frictional stress tensor which depends on the space derivatives of the velocity, i.e. the strain tensor. ii) The relationship between F ij and e ij is linear. iii) The relationship does not depend on the frame of reference. We have already shown that, with reference to the respective principal axes, T & : 0 0 ' & e 0 0 ' 0 : 0 and * 0 e. (5.44) 0 " 0 0 : # " 0 0 e # Because :,, and are normal stresses and e, e, e : : are normal strains, it is natural to assume that the principal axes of both tensors are the same. We therefore assume that : p ) V + *e 5 : p ) V + *e, : p ) V + *e 4 (5.45) where p satisfies (5.9). Writing direction cosines between the principal axes (X, X, X ) and axes of a Cartesian coordinate system (x, y, z) as shown in Table 5.,

13 5.6 Physical laws connecting the stress and strain tensors 05 x y z a b g X X a b g X a b g Table 5.: Direction cosines between principle axes. we find that : xx : + : + : : xy 0 : + 0 : + 0 :, (5.46) etc. Substituting (5.45) into (5.46), and using (5.4), we obtain & : xx & p ) V# + *e ' + & p ) V# + *e ' + & p ) V# + *e ' p + *e xx, : xy & p ) V# + *e ' 0 + & p ) V# + *e ' 0 " + & p ) V# + *e ' 0 *e xy, (5.47) etc. In general we have : ij p! ij ) V! ij + *e ij. (5.48) This expression gives : ii p ) V + *. ) V p + + * #. ) V. (5.49) Because we have defined : ij p, we conclude + * + 0. (5.50) Thus we have the final expression

14 06 : ij p! ij + * & e ij --. ) V! ' " ij # p! ij + F ij. (5.5) 5.7 The Navier-Stokes equation Substitution of (5.5) into (5.4) gives the resultant forces per unit volume, as expressed by the divergence of the stress tensor: DivT.p + DivF (5.5) The equation of motion in the inertial frame (with subscript a omitted) is DV A.p + DivF Dt (5.5) or, in Cartesian tensor notation, Dv i Dt (v i ( v v i (t j (x j (A ( p ( C D * e (x i ( x i ( x ij -- ". ) V#! j ij 4 5 B (A ( p ( ( * v i (v j (x i ( x i ( x " & j (x j (x # ' ( --* v k C D B i (x k (5.54) This is called the Navier-Stokes equation. When the fluid is incompressible ". ) V 0#, and * is spatially constant, (5.54) reduces to Dv i Dt (A ( p * ( v i & ' +, (x ( x i " # (x j (5.55) or, in vector form, DV A.p + *. V. Dt (5.56) Here we see the Laplacian of the vector V.

15 5.8 A proof that the dissipation is non-negative 07 (v It should be noted that e ij -- i (v & j ' vanishes when V r (rigid " (x j (x # i rotation), and that. ) " r# 0. This means that we do not have to worry about the distinction between V a V + r and V in the expression F DivF in the equation of relative motion. The frictional force is the same in the rotating frame as in the inertial frame. 5.8 A proof that the dissipation is non-negative The dissipation rate is given by ( v " F ).# ) V F i ij ! (x j + + F F F ( u (x F ( u (y F ( u (z ( v (x ---- F ( v (y ---- F ( v (z ( w (x F ( w (y F ( w. (z (5.57) Since F ij F ji, we have ( v! F i ij (x F ( v i ji j (x F ( v j, ij j (x i (5.58) so we can write ( v! F i (v ij j " & (x j (x # '. i (5.59) But F ij * -- (v i (v j " & (x j (x # ' ( -- v i ! i (x ij i * -- (v i (v j " & (x j (x # ' (v "!ij # -- i (v j i " & (x j (x # ' ( -- v i C D ! i (x ij 4 5, B i (5.60) or

16 08 F ij * -- (v i (v j " & (x j (x # ' ( "! ij # -- v i ! i (x ij. j (5.6) Substituting (5.6) into (5.59), we find that! * -- (v i (v j " & (x j (x # ' ( "!ij # -- v i ! i (x ij j &(v i (v j ' "(x j (x # i * e ij e ij "! ij # ". ) V#! ij. (5.6) This expression is obviously non-negative, since "! ij # E 0 and! ij E 0. For a nondivergent flow, the dissipation can be expanded, using Cartesian coordinates, as! *e ij e ij --* (u (v " & (y (x# ' ( u (w " & (z (x# ' (v (u " & (x (y# ' ( v (w " & (z (y# ' (w (u " & (x (z# ' ( w (v " & (z (z# ' * (u (v " & (y (x# ' ( u (w " & (z (x# ' ( w (v " & (y (z# '. (5.6) This shows that! is the sum of squares; again, it cannot be negative.