UNIVERSITY OF EAST ANGLIA School of Mathematics Main Seies UG Examination 2015 16 FLUID DYNAMICS WITH ADVANCED TOPICS MTH-MD59 Time allowed: 3 Hous Attempt QUESTIONS 1 and 2, and THREE othe questions. Notes ae not pemitted in this examination. Do not tun ove until you ae told to do so by the Invigilato. MTH-MD59 Module Contact: D Richad Puvis, MTH Copyight of the Univesity of East Anglia Vesion: 1
- 2 - Note: Below ae the Navie-Stokes equations, fo an incompessible fluid of constant density ρ and constant kinematic viscosity ν in the absence of gavity, in which the velocity u = u e + u θ e θ + u z k and the pessue is p, expessed in cylindical pola coodinates, θ, z : u = 1 (u ) + 1 u θ θ + u z z = 0, u θ + u u z z u2 θ = 1 ρ u t + u u + u θ u θ t + u u θ + u θ u z t + u u θ θ + u z u z + u θ whee the Laplace opeato is 2 f = 1 u θ z + u u θ u z u z θ + u z ( p + ν 2 u u 2 ) u θ, 2 2 θ = 1 ( p ρ θ + ν 2 u θ u θ + 2 ) u, 2 2 θ z = 1 p ρ z + ν 2 u z, ( f ) + 1 2 f 2 θ + 2 f 2 z. 2 In spheical pola coodinates, θ, α, with u = u e + u θ e θ + u α e α, F = F e + 1 F θ e θ + 1 F sin θ α e α, and u = 1 2 ( 2 u ) + 1 sin θ θ (u θ sin θ) + 1 u α sin θ α = 0. MTH-MD59 Vesion: 1
- 3-1. (i) (a) Deive fom fist pinciples the continuity equation fo the flow of an incompessible fluid though a poous medium of void faction φ(x, t). Assuming that Dacy s law holds fo gavity diven flow, namely u = k (p + ρgy), µ and assuming that both φ and the pemeability k ae constant, show that 2 p = 0. [6 maks] (b) Biefly explain the assumptions which pemit the bounday conditions (p + ρgz) = 0 n at an impemeable fixed bounday, whee n is the diection nomal to the bounday, and p = 0, and f t + (u ) f = 0, at a bounday f(x, t) = 0 which sepaates a satuated egion fom a dy one which is open to the atmosphee. [2 maks] (ii) In a poous medium, wate flows in a shallow laye of thickness h(x, t) above an impemeable base y = 0. The poous medium above the wate laye is dy and open to the atmosphee. (a) Justify the appoximate model u = (u, v) whee u = k p µ x, 0 = p y + ρg, u x + v y = 0, with v = 0 on y = 0, and p = 0, v h t uh x = 0 on y = h. [3 maks] (b) Hence show that µ h kρg t = ( h h ). [5 maks] x x (c) Show that this equation can have a similaity solution of the fom h(x, t) = f(η), whee η = xt α, fo a value of α that should be found. Also find the odinay diffeential equation satisfied by f, but do not attempt to solve it. [4 maks] MTH-MD59 PLEASE TURN OVER Vesion: 1
- 4-2. A viscous incompessible fluid flows steadily down the inside of a vetical pipe of cicula coss-section of adius a. The flow is due to gavity (of constant acceleation gk ), and a given constant pessue gadient dp/dz, whee the positive z axis points vetically up. Use cylindical pola coodinates, θ, z, whee = a is the fixed pipe bounday. (i) Show fom the equations of motion that it is consistent to assume that the velocity field u is paallel to k, and that u vaies with espect to only. State the appopiate bounday conditions. [4 maks] (ii) Consequently show that u = 1 4 c ( a 2 2) k,. whee the constant c should be found. (iii) Find the volume flux of fluid flowing though the pipe. [10 maks] [3 maks] (iv) Suppose we additionally have an inne concentic stationay cylinde of adius = b, b < a. Find the velocity in this case, and explain why you would expect the flux to be lowe, even if b 1. (Thee is no need to explicitly calculate the flux in this case). [3 maks] MTH-MD59 Vesion: 1
- 5-3. Ai flows above a fixed impemeable plate located at y = 0 fo x 0. The flow has a high Reynolds numbe. (i) (a) Stating fom the steady two-dimensional non-dimensionalised Navie-Stokes equations in (x, y ), explain why fo a high Reynolds numbe flow we expect a bounday nea the wall. (b) By escaling the nomal coodinate as y = Re 1/2 Y, and v appopiately, show that the equations educe to the bounday laye equations uu x + V u Y = U du dx + u Y Y, u x + V Y = 0 whee in the fa-field as y, u U (x)i and p x U du dx. (ii) The fa-field flow has the fom U = x 2. [8 maks] (a) By intoducting the steamfunction ψ(x, Y ) such that u = ψ Y and V = ψ x, show that ψ must be a solution of the patial diffeential equation ψ Y ψ xy ψ x ψ Y Y = 2x 3 + ψ Y Y Y. [3 maks] (b) Given that we can wite ψ = x 3 2 f(η) with η = Y x 1 2, find the velocity components u and v in tems of x, y, f and df dη. (c) Hence show that f(η) must satisfy the odinay diffeential equation ( ( ) ) d 3 2 f dη + Af d2 f df 3 dη + B 1 = 0, 2 dη [4 maks] fo constants A and B that should be found, and find the bounday conditions that f(η) must satisfy. [5 maks] MTH-MD59 PLEASE TURN OVER Vesion: 1
- 6-4. An inviscid fluid flows steadily without gavity between impemeable walls at y = 0 and y = 1. The velocity is U(y) and the pessue is zeo. This steady flow is subjected to small time-dependent distubances, such that the distubed flow velocity and pessue can be witten as u(x, y, t) = ( U(y) + û(y)e i(kx ωt)) i + ˆv(y)e i(kx ωt) j, p(x, y, t) = ˆp(y)e i(kx ωt), whee the eal pat of the ight-hand sides is undestood. The distubances û, ˆv and ˆp ae assumed to be small. (i) Explain biefly how looking fo such solutions tells us about the stability of the flow. [4 maks] (ii) Stating fom Eule s equations fo inviscid flow, along with u = 0 deive the thee diffeential equations that goven û, ˆv and ˆp. You may neglect small tems such as û 2, ˆv 2, ˆvˆv and should find equations that ae linea. [6 maks] (iii) Eliminate û and ˆp to show that ˆv(y) satisfies d 2ˆv ( k dy + d 2 ) U 2 ω Uk dy 2 k2 ˆv = 0, subject to ˆv = 0 at y = 0 and ˆv = 0 at y = 1. [6 maks] (iv) Let that the eal constant k > 0, and wite ω = ω R + iω I. Fo a geneal U(y), a consequence of ou equation fo ˆv is 1 d 2 U ˆv 2 dy ω I k 2 2 dy = 0. 0 ω Uk Use this to deduce that a necessay condition fo the flow to become unstable is that d2 U dy 2 must change sign in 0 < y < 1. [4 maks] MTH-MD59 Vesion: 1
- 7-5. (i) Descibe biefly what is meant by the lubication appoximation to the Navie- Stokes equations. Unde what conditions is it valid? [2 maks] (ii) The lubication appoximation fo the gavity-diven flow in a thin liquid film sitting on a solid suface at y = 0, with a fee-suface at y = h(x, t) educes the Navie-Stokes equations to 0 = p x + µ 2 u y 2, 0 = p y 1, u x + v y = 0, in non-dimensional fom. Assuming that the stess-fee bounday conditions on the fee-suface ae that p = 0 and u y = 0 on y = h, state the appopiate conditions on y = 0 and show that the velocity can be witten as u(y) = 1 h y(2h y). 2µ x [5 maks] (iii) By integating the continuity equation, o othewise, show that the flow is govened by h t = 1 3µ x ( h 3 h ). x [You may assume that the kinematic condition equies v = h + u h t x on the feesuface at y = h(x, t).] (iv) Suppose that the fee-suface can be witten in the fom h = h 0 + η(x, t), [6 maks] and η(x, t = 0) = η 0 exp ( x 2 ) whee h 0 and η 0 ae constants with η 0 h 0. Show that η is govened to leading ode by and show that the solution given by η(x, t) = η t = 1 η 0 (1 + 4Dt) 1 2 3µ h3 0 2 η x, 2 } exp { x2, 1 + 4Dt satisfies this equation and the bounday conditions fo a paticula value of D which should be found. [7 maks] MTH-MD59 PLEASE TURN OVER Vesion: 1
- 8-6. (i) State the Stokes equations fo incompessible slow viscous flow and show that they educe to u = 0. (1) You may use the vecto identity 2 u = ( u) u without poof. [6 maks] (ii) An axisymmetic incompessible Stokes flow has the Stokes steamfunction ψ(, θ) in spheical pola coodinates (, θ, α), defined by the velocity components u = 1 ψ 2 sin θ θ, u θ = 1 ψ sin θ. Fluid is contained between two sphees at = 1 and = b, with b > 1. The fluid velocity vanishes on the oute sphee, while on the inne sphee u = U cos θ and u θ = 0. (a) You ae given that the geneal solution to the govening equation (1) has the fom ( ) A ψ(, θ) = U sin 2 θ + B + C2 + D 4. Find the components of velocity and wite down the conditions elating the constants A, B, C and D that allow all the bounday conditions to be satisfied. You do not need to solve to find the constants. [6 maks] (b) In the limit as b, with u 0 as, show that A = B = 1/4 with C = D = 0. [4 maks] (c) In the limit of b vey lage but finite, the constants can be witten in the appoximate fom A 1 4 + a 1 b, B 1 4 + b 1 b, C c 1 b, D d 1 b 3. ignoing highe-ode tems. Find the constants a 1, b 1, c 1 and d 1. [4 maks] END OF PAPER MTH-MD59 Vesion: 1