An Exact Solution of Navier Stokes Equation

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1 An Exact Solution of Navie Stokes Equation A. Salih Depatment of Aeospace Engineeing Indian Institute of Space Science and Technology, Thiuvananthapuam, Keala, India. July 20 The pincipal difficulty in solving the Navie Stokes equations (a set of nonlinea patial diffeential equations) aises fom the pesence of the nonlinea convective tem (V )V. Since thee ae no geneal analytical methods fo solving nonlinea patial diffeential equations exist, each poblem must be consideed individually. Fo most pactical flow poblems, convective acceleation of fluid paticles cannot be ignoed. Howeve, in geneal, exact solutions ae possible only when the nonlinea tems vanishes identically. Thee ae a few special cases fo which the convective acceleation vanish because of the natue of the geomety of the flow system. In these cases exact solutions ae usually possible and below we conside one of such poblems. Pipe Flow Induced by Movement of Wall Hee we conside an infinitely long hoizontal cicula pipe filled with a Newtonian fluid of density ρ and viscosity µ. Fo t < 0 both pipe and fluid ae at est. At time t = 0, the pipe impulsively stats moving in the axial diection with a unifom velocity U. As a consequence of this, the fluid movement is induced in the axial diection as sketched in figue. The momentum equation fo incompessible flow in cylindical coodinate system Figue : Fluid flow in a pipe induced by the motion of pipe walls.

2 is given by ( Du ρ Dt ( Duθ ρ Dt u2 θ + u u θ ) ρ Du z Dt = ρg p ( + µ ) = ρg θ p θ + µ = ρg z p z + µ 2 u z 2 u 2 u θ 2 θ u ) ( 2 2 u θ + 2 u 2 θ u ) θ 2 Fo the conditions of the pesent unsteady paallel flows in the absence of body foces, many tems disappea and the axial velocity u z is the only nonzeo velocity component. Theefoe the fist two equations become tivial and the last equation (axial momentum equation) educes to ρ u z t [ = µ ( u ) z + 2 ] u z 2 θ 2 Since the flow is otationally symmetic, equation () educes to u z = ν ( u ) z t The initial and bounday conditions ae u z (, 0) = 0 (initial condition) u z (0, t) is bounded (3) u z (R, t) = U (no-slip condition) Fo convenience, we intoduce the following dimensionless quantities: u = U u z U η = R τ = νt R 2 Thus the diffeential equation (2) can be witten in dimensionless quantities as follows: ν U u R 2 τ = ν U [ 2 u R 2 η 2 + ] u η η u τ = 2 u η 2 + u η η Coesponding dimensionless initial and bounday conditions ae u(η, 0) = (initial condition) u(0, τ) is bounded (5) u(, τ) = 0 (no-slip condition) () (2) (4) 2

3 Note that the new bounday conditions ae homogeneous. Although the poblem defined by (4) and (5) is time dependent, it is linea in u and confined to the bounded spatial domain, 0 η. Thus it can be solved by the method of sepaation of vaiables. In this method we fist find a set of eigensolutions that satisfy the diffeential equation (5) and the bounday condition at η = 0 and η = ; then we detemine the paticula sum of those eigensolutions that also satisfies the initial condition at τ = 0. The poblem (4) and (5) compises one example of the geneal class of so-called Stum Liouville poblems fo which an extensive theoy is available that ensues the existence and uniqueness of solutions constucted by means of eigenfunction expansions by the method of sepaation of vaiables. We begin with the basic hypothesis that a solution of (5) exists in the sepaable fom and choose the following ansatz: u(η, τ) = F(η)G(τ) (6) Substituting this ansatz into equation (4) to obtain F dg dτ = F Gd2 dη 2 + G df η dη (7) As G depends only on τ and F only on η, by sepaation of vaiables the following odinay diffeential equations fo G and F esult: dg G dτ = λ 2 (8) d 2 F F dη 2 + df F η dη = λ 2 (9) Hence the oiginal poblem govened by a PDE has now been tansfomed into two elated auxiliay poblems govened by ODEs. The choice of a negative constant λ 2 is due to the fact that the solution will decay to zeo as time inceases, i.e., u 0 as τ. The solution fo the diffeential equation (8) can be deived by integation: G = c 0 e λ 2 τ (0) whee c 0 is an integation constant to be detemined. In ode to detemine the solution of the diffeential equation fo F(η), equation (9) can be witten as follows: d 2 F dη 2 + df η dη + λ 2 F = 0 o η 2 d2 F dη 2 + η df dη + λ 2 η 2 F = 0 () 3

4 o, on intoducing a change of independent vaiables, y = λη we obtain y 2 d2 F dy 2 + ydf dy + y2 F = 0 (2) This is Bessel diffeential equation of ode 0. It has two linealy independent solutions, J 0 (y) and Y 0 (y) which ae known as Bessel functions of the fist and second kinds of ode 0. Thus the most geneal solution of equation (2) can be witten as F(y) = c J 0 (y) + c 2 Y 0 (y) (3) A plot showing the behavio of these two functions is shown in figue (2). Both oscillate Figue 2: The Bessel functions of the fist and second kinds of ode 0. back and foth acoss zeo, but Y 0 (y) as y 0. Substituting equations (0) and (3) into (6), we obtain the geneal solution of the fom u(y, τ) = e λ 2τ [c J 0 (y) + c 2 Y 0 (y)] (4) Note that the constant c 0 has been dopped since it is edundant hee. The solution is bounded at y = 0. This condition is satisfied only when the constant c 2 = 0. Hence, the geneal solution (4) (that is bounded at y = 0) takes a fom: u(y, τ) = c J 0 (y)e λ 2 τ (5) 4

5 Theefoe, the solution u, afte substituting y = λη, may be witten as u n (η, τ) = A n J 0 (λ n η)e λ 2 n τ (6) The subscipt n is added in anticipation of the fact that thee is an infinite, but discete, set of values possible fo λ such that the geneal solution (5), satisfies the bounday condition u = 0 at η =. This set of values of λ = λ n is known as the eigenvalues of the poblem, and the coesponding u n ae called the eigenfunctions. To detemine the eigenvalues λ n, we apply the bounday condition at η = to equation (6), that is, 0 = A n J 0 (λ n )e λ 2 n τ fo all Since setting A n = 0 would esults in a tivial solution, one must equie τ J 0 (λ n ) = 0 (7) fo the non-tivial solution. Theefoe, one obtains multiple values of λ n (eigenvalues) that satisfy the bounday conditions at the wall. Clealy these eigenvalues ae equal to the infinite set of zeoes of the Bessel function of ode zeo, J 0 (z). Refeing to figue 2, we have denoted those zeoes as s n, with the fist cossing fo the smallest value of z being s, namely, and thei values ae obtained as λ n = s n, n =, 2, 3,..., (8) s n = 2.405, 5.520, 8.654,.792, 4.93, 8.07, 2.22, , (9) Each of the solutions of λ n now constitutes an individual solution. Consideing the lineaity of the govening equation and bounday conditions (4) and (5), the complete solution fo u n (η, τ) is obtained by linea supeposition: u = u n (η, τ) = A n e λ n 2τ J 0 (λ n η) (20) whee A n ae abitay, constant coefficients. Equation (20) is called the Fouie Bessel seies. This solution satisfies the diffeential equation (4) and the bounday condition u = 0 at η = fo any choice of the constant coefficients A n. The final step is to choose the A n so that u(η, τ) satisfies the initial condition u(η, 0) =. The geneal Stum Liouville theoy guaantees that the eigenfunctions (6) fom a complete set of othogonal functions. Thus it is possible to expess the smooth initial condition by means of the Fouie Bessel seies (20) with τ = 0, that is, u(η, 0) = A n J 0 (λ n η) = A J 0 (λ η) + A 2 J 0 (λ 2 η) +... A n J 0 (λ n η) +... = (2) 5

6 To detemine the A n, we will take advantage of othogonality popeties of J 0 : J 0 (λ m η)j 0 (λ n η)η dη = 2 J 2(λ n) if n = m 0 0 if n m whee J is the Bessel function of fist kind of ode. Multiplying both sides of (2) by ηj 0 (λ m η) and integating ove η fom 0 to, we obtain A n 0 η J 0 (λ m η)j 0 (λ n η)dη = 0 η J 0 (λ m η)dη Due to the othogonality popety (22), the only nonzeo tem on the left hand side is that fo m = n hence, 0 η J 0 (λ n η)dη 0 A n = 0 η J0 2(λ nη)dη = η J 0 (λ n η)dη 2 J2 (λ (23) n) Fo evaluating the numeato of (23) we make use of the following popety of Bessel function: Theefoe (22) η p+ J p (λη)dη = λ η p+ J p+ (λη) (24) η J 0 (λ n η)dη = λ n ηj (λ n η) (25) and thus, the numeato of equation (23) becomes η J 0 (λ n η)dη = J (λ n ) (26) λ n 0 Substitution of equation (26) into (23) yields an expession fo A n : λ A n = n J (λ n ) 2 J2 (λ n) = 2 [J (λ n )] λ n Thus fo the velocity distibution accoding to (20), the following expession esults: u(η, τ) = 2 (27) λ n J (λ n ) e λ 2 n τ J 0 (λ n η) (28) The above Fouie Bessel seies has the popety of conveging vey quickly when the dimensionless time τ = νt/r 2 is lage. On the othe hand, the convegence is slow when τ is small. Reveting to dimensional vaiables, we can expess the solution of the full, oiginal poblem in tems of the axial velocity pofile: [ u z (, t) = U 2 ( λ n J (λ n ) e λ n 2 νt/r 2 ) ] J 0 λ n (29) R Obviously, as t, this solution evets to the steady-state unifom flow pofile. obtain othe details of this velocity pofile, it is necessay to evaluate the infinite seies numeically fo each value of t and. A typical numeical example of the esults is shown in figue 3, whee u z has been plotted vesus fo seveal values of t. 6 To

7 Figue 3: Tansient velocity pofiles in a pipe induced by the motion of the pipe walls. Shea stess distibution The shea stess distibution in the flowfield can be easily obtained fom the velocity distibution as follows In dimensionless fom whee dimensionless shea stess τ x = ( u τ x = µ x + u ) z = µ du z d τ x = du dη τ x µu/r. pofile, we make use of the following popety of Bessel function: Theefoe and thus the shea stess pofile is given by Fo evaluating the deivative of velocity d [ η p J p (λη) ] = λη p J p+ (λη) (30) dη d dη [J 0(λη)] = λj (λη) (3) τ x = du dη = 2 [J (λ n )] e λ n 2τ J (λ n η) (32) Fo the value of wall shea stess at the pipe wall, i.e. at η =, one obtains τ wall = τ x η= = du dη = 2 e λ n 2 τ η= (33) Reveting to dimensional vaiables, we can expess the wall shea stess in tems of oiginal vaiables: τ wall = du z d = 2µU =R R e λ n 2 νt/r 2 (34) 7

8 It can be seen that the wall shea stess has a finite value, even fo time t = 0. This is a supising esult when compaing to othe cases whee flow is induced by impulsively motion of the bounday. Note also that the wall shea stess τ wall 0 as t. This is obvious because the steady-state velocity pofile is unifom. 8

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