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The Navier-Stokes Equation

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The Navier-Stokes Equation Revised October 9, 2007 8:48 Copyright 2000 David A. Randall The Navier-Stokes Equation David A. Randall Department of Atmospheric Science Colorado State University, Fort Collins, Colorado 80523 25.1 Analysis of the relative motion near a point Suppose that the velocity of the fluid at position r0"x! y! z# and time t is V"x! y! z! t# , and that the simultaneous velocity at a neighboring position r0 + r is V + !V . We can write & u u u ' $ -(---- -(---- -(---- % (x (y (z & !u' $ % &x' $ v v v % $ !v% = $ -(---- -(---- -(---- % $y% . (25.1) $ % $ (x (z (z % $ % "!w# "z# $ w w w % $ -(----- -(----- -(----- % " (x (y (z # In vector form, this can be written as !V = D ) r , (25.2) 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 (25.1) as (vi !vi = ------xj , where i = 1! 2! 3 . (25.3) (xj We can decompose (vi (xj into two parts which are symmetric and anti- symmetric in the suffices i and j . They are 1 (v (v * = --&------i + ------j' , (25.4) ij " # 2 (xj (xi and Selected Nuggets 193 194 The Navier-Stokes Equation 1 (v (v + = --&------i – ------j' , (25.5) ij " # 2 (xj (xi respectively. The deformation D can now be expressed by D = * + , , (25.6) where we define two “parts” of D : & e e e ' & 0 + + ' $ xx xy xz % $ xy xz % * - $ e e e % , and , - $ + 0 + % . (25.7) $ yx yy yz % $ yx yz % " ezx ezy ezz # " +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. "s# "a# "s# "a# !V = !V + !V , where !Vi = eijXj , and !Vi = +ijXj . (25.8) 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 (u (v (w diagonal elements of * , i.e. e = -----! e = -----! e = ------ , are called the normal strains, xx (x yy (y zz (z and represent the rate of volume expansion, as illustrated in Fig. 25.1 a. The sum of the normal strains, i.e., the trace of the strain tensor, is the divergence of V , i.e. exx + eyy + ezz = eii = . ) V . (25.9) The off-diagonal elements of * , namely exy , eyz , and ezx , are called the shearing strains, and express the rate of shearing deformation, as illustrated in Fig. 25.1 b. They can be written as 1 (u (v e = e = --&----- + -----' $, xy yx 2"(y (x# 1 (v (w e = e = --&----- + ------' $, yz zy 2"(z (y# 1&(w (u' ezx = exz = -- ------ + ----- $. 2" (x (z# (25.10) Selected Notes of David A. Randall 25.1 Analysis of the relative motion near a point 195 y a dv u v -(---- + -(---- (x (y Dy x 0 Dx du y b du u v -(---- + -(---- (y (x Dy d v x 0 Dx Figure 25.1: a) The normal strains, representing the rate of volume expansion. b) The shearing strains, which represent the rate of shearing deformation. The elements of the anti-symmetric tensor, , , are 1 (w (v 1 (u (w (v (u + = –+ = --&------ – -----'! + = –+ = --&----- – ------'! + = –+ = &----- – -----'$ . (25.11) zy yz 2" (y (z# xz zx 2" (z (x# yx xy "(x (y# Introducing the notation /x = –2+yz! /y = –2+zx! /z = –2+zx , we see that "/x! /y! /z# is the vorticity vector / = . % V$ . (25.12) Selected Notes of David A. Randall 196 The Navier-Stokes Equation In tensor notation, (25.12) can be written as (vk /i = *ijk------- , (25.13) (xj where *ijk = 1 when i, j, k are in even permutation, and *ijk = –1 when i, j, k are in odd (v2 (v1 (v (u permutation. For example, /z = /3 = *321------- + *312------- = ----- – ----- , which gives (x1 (x2 (x (y twice the angular velocity about the z-axis, as illustrated in Fig. 25.2. Thus -du Dy (v (u 0˙ ----- – ----- = 1˙ + 0˙ (x (y dv 1˙ Dx Figure 25.2: The component of the vorticity in the z direction. & "a# ' & ' & ' $ !u % 0 –/z –/y x $ % 1$ % $ % "a# = --$ / 0 –/ % $ y % , (25.14) $ !v % 2$ z x % $ % $ "a# % z " !w # " –/y /x 0 # " # or "a# 1 !V = --"/ % r# . (25.15) 2 25.2 Invariants of the strain tensor Consider the scalar product Selected Notes of David A. Randall 25.2 Invariants of the strain tensor 197 r ) * ) r = r ) "* ) r# 2 2 2 = exxx + eyyy + ezzz + 2"exyxy + eyzyz + ezxzx# = f"x! y! z#$. (25.16) By putting f"x! y! z# = constant , we obtain a surface of second order, the strain quadratic, also called the ellipsoid of strain. We refer the strain quadratic to its principal axes "X1! X2! X3# . Then (25.16) takes the form 2 2 2 e1X1 + e2X2 + e3X3 = F"X1! X2! X3# = constant . (25.17) Here e1 , e2 , and e3 are the principal values of the strain tensor, and are called the principal normal strains. Using (25.16), we see that "s# 1 (f 5 !u = ------- = e x + e y + e z 2 x xx xy xz 3 ( 3 "s# 1 (f 3 !v = ------- = eyxx + eyyy + eyzz 4$ . (25.18) 2(y 3 "s# 1 (f 3 !w = ------ = ezxx + ezyy + ezzz 3 2(z 2 In terms of the principal axes "X1! X2! X3# , "s# 1 (F "s# "s# !v1 = ---------- = e1X1! !v2 = e2X2! !v3 = e3X3 . (25.19) 2(X1 "s# "s# "s# "s# "!v1 ! !v2 ! !v3 # are the projections of !v 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. exx + eyy + ezz = eii = . ) V . (25.20) This quantity is an invariant under the rotation of the axes of reference. Consider the Selected Notes of David A. Randall 198 The Navier-Stokes Equation transformation by which the surface f"x! y! z# = the constant of (25.16) is transformed into F"X1! X2! X3# = the constant of (25.17). We can write the transformation between the two Cartesian coordinates in the schematic, tabular form shown in Table 25.1 x y z X1 a1 b1 g1 X2 a2 b2 g2 X3 a3 b3 g3 Table 25.1: 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 3 3 2 5 1i = 1! 1i0i = 0 3 6 6 $ , (25.21) i = 1 i = 1 4 3 1i1j + 0i0j + 7 i7 j = !ij 2 where !ij is Kronecker’s delta "!ij = 1 when i$ $= j! !ij = 0 when i$ 8$ j# . Substituting the expression Xi = 1ix + 0iy + 7 iz where i = 1! 2! 3 (25.22) into (25.17), we obtain 3 3 2 5 1i = 1! 1i0i = 0 3 6 6 . (25.23) i = 1 i = 1 4$ 3 1i1j + 0i0j + 7 i7 j = !ij 2 This expression must be identical with (25.16), so that 2 2 2 5 exx = 11e1 + 12e2 + 13e3! exy = 1101e1 + 1202e2 + 1303e3 , 3$ $ 3 2 2 2 . (25.24) eyy = 01e1 + 02e2 + 03e3! eyz = 017 1e1 + 027 2e2 + 037 3e3 , $ 4 $$ 3 e = 2e + 2e + 2e e = e + e + e , 3 zz 7 1 1 7 2 2 7 3 3! zy 7 111 1 7 212 2 7 313 3 $2 Selected Notes of David A. Randall 25.3 Stress 199 It follows that 3 2 2 2 9 = exx + eyy + ezz = 6 "1i + 0i + 7 i #ei = e1 + e2 + e3 (25.25) i = 1 i.e., the sum exx + eyy + ezz = . ) V is independent of the choice of reference. Of course, this quantity is called the divergence, and we “knew already” that it is independent of the choice of reference. Another invariant of the strain tensor is 2 2 2 exxeyy + eyyezz + ezzexx – exy – eyz – ezx = constant . (25.26) The proof is left as an exercise. 25.3 Stress In a moving viscous fluid, forces act not only normal to a surface but also tangential to it. In Fig. 25.3, :x , :y and :z denote respectively the forces per unit area acting upon the surfaces normal to the positive x, y, and z directions, respectively. In z (: : + -------z9z z (z :–x = –:x (: : + -------y9y y (y 0 y :–y = –:y (: x : + -------x9x : = –: x (x –z z Figure 25.3: Sketch illustrating the normal and tangential forces acting on an element of fluid. Selected Notes of David A. Randall 200 The Navier-Stokes Equation general, each force is a vector with 3 components, so that :x = :xxi + :yxj + :zxk , 5$ $ 3 :y = :xyi + :yyj + :zyk , 4$ $ . (25.27) 3 :z = :xzi + :yzj + :zzk . $2 Symbolically we can write an element of the stresses as :ij"i = 1! 2! 3;j = 1! 2! 3# . Here :ij 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 & : : : ' $ xx xy xz % T = $ : : : % (25.28) $ yx yy yz % " :zx :zy :zz # For an arbitrarily chosen plane surface, whose normal vector is n , the force acting per unit area is :n . See Fig. 25.4. The equilibrium condition is :–y = –:y (25.29) :n9; – :x9;x – :y9;y – :z9;z = 0 , 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 .
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