Thin flow over a sphere

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Winfred Chandler
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1 1 University of Waterloo, Faculty of Mathematics, Waterloo, Canada 2 Ryerson University, Department of Mathematics, Toronto, Canada IMA8 2016, June 12-16, Bad Honnef, Germany

2 Problem description Related problem Previous work Governing equations Scaling Dimensionless equations Boundary conditions Small δ expansion Leading-order problem O(δ) problem

Problem description Related problem Previous work The steady flow of a thin fluid layer over a sphere occurs naturally in a globe fountain as shown.

3 Problem description Related problem Previous work The steady flow of a thin fluid layer over a sphere occurs naturally in a globe fountain as shown. A globe fountain consists of a sphere with a hole at the top whereby water is pumped out at a constant rate. An analytical study of the flow has been carried out in this study.

A striking example of levitation is encountered in the kugel fountain where a granite sphere, Outline sometimes weighing over a tonne, is kept aloft by a thin film of flowing water.

4 A striking example of levitation is encountered in the kugel fountain where a granite sphere, Outline sometimes weighing over a tonne, is kept aloft by a thin film of flowing water. In this paper we Problem description explain the working principle Mathematical behind this formulation levitation. We show that the fountain can be viewed as Related problem a giant roller bearing and thus Analytical forms a solution prime example Previous of lubrication work theory. It is demonstrated how the viscosity and flow rate of the fluid determine (i) the remarkably small thickness of the film supporting the sphere and (ii) the surprisingly long time it takes for rotations to damp out. The theoretical results compare well with measurements on a fountain holding a granite sphere of one meter in diameter. We close by discussing several related cases of levitation by lubrication. I. INTRODUCTION ite sphere or kugel fountains (see Fig. 1) are The kugel shown here iar sight in town squares and science museums, aller ones involves often with a massive a marble granite sphere decorate rivate homes and gardens. These fountains conperfectly polished ball floating in a socket that sphere which floats and spins cisely around on a thin it. The filmfluid ofthat flowing wells up water around of the socket is pumped into the fountain via a which is out of a hole the base. In spite of its considerable weight, the is easily atbrought the base into of a spinning the fountain. motion, which tractive sight especially when the surface of the is engraved with the Earth s map, a soccer ball, ht sky, or something of the kind. The fluid layer the socket and sphere is very thin (thinner than card 1 ), which is important for any kugel on disa public place, since it means there is no risk of s fingers being caught under the spinning sphere. ite its popularity, the granite sphere fountain is understood by most people. When we asked vis- FIG. 1: One of the largest granite sphere fountains i world, the Grand Kugel at the Science Museum of Vir Richmond. The sphere has a diameter of 2.65 m and a of about 27 tonnes.

5 Problem description Related problem Previous work Takagi & Huppert (JFM, 2010) studied a constant volume of fluid released at the top of a sphere using lubrication theory. Their analytical results agreed well with their experiments. They also investigated the onset of instability of the advancing front as it split into a series of rivulets. The dynamics of the kugel fountain was recently analysed by Snoeijer & van der Weele (Am. J. Phys., 2014), again using lubrication theory.

6 Governing equations Scaling Dimensionless equations Boundary conditions Formulated in spherical coordinates (r, θ, φ) with the hole oriented about the polar axis θ = 0 and assuming azimuthal symmetry, the governing steady-state Navier-Stokes equations become: + ν r 2 [ r + ν r 2 [ r ( r 2 v r ( r 2 v sinθ ) + (ru sinθ) = 0, r θ v v r + u r ) + 1 sinθ v u r + u r ( r 2 u r v θ u2 r θ u θ + uv r ) + 1 sinθ = 1 P ρ ( sinθ v θ θ r g cosθ ) 2 sinθ = 1 P ρr θ + g sinθ ( sinθ u ) + 2 v θ θ (u sinθ) 2v θ u ] sin 2 θ. ],

7 Governing equations Scaling Dimensionless equations Boundary conditions Here, u, v denote the velocity components in the θ and radial directions, respectively, P refers to the pressure, ν = µ/ρ represents the kinematic viscosity while µ is the dynamic viscosity and ρ is the fluid density whereas g is the acceleration due to gravity. The thickness of the fluid layer is scaled by the Nusselt thickness, H, which for a vertical incline is given by H 3 = 3νQ g, where Q is the constant flow rate exiting the small hole. For a fluid layer having a width of unity Q = UH where U is the velocity scale.

8 Governing equations Scaling Dimensionless equations Boundary conditions The coordinate y is introduced which is related to r through the relation r = R + y with R referring to the radius of the sphere. This can be scaled as r ( y ) R = 1 + δ, H where the dimensionless parameter δ = H/R 1 denotes the shallowness parameter. With this scaling the dimensionless flow variables and coordinate y are given by (u, v, P, y) (Uũ, δuṽ, ρu 2 P, Hỹ), where the tilde denotes a dimensionless quantity.

9 Governing equations Scaling Dimensionless equations Boundary conditions In dimensionless form and suppressing the tildes the governing equations become: [ (1 + δy) 2 v sinθ ] + [(1 + δy)u sinθ] = 0, y θ δ 2 v v y + δ2 u v (1 + δy) θ δu2 (1 + δy) = P y 3 cosθ δ + Re Re(1 + δy) 2 [ ( (1 + δy) 2 v ) ( + δ2 sinθ v ) 2δ ] y y sinθ θ θ sinθ θ (u sinθ) 2δ2 v δv u y + δu u (1 + δy) θ + δ2 uv (1 + δy) = δ (1 + δy) [ ( (1 + δy) 2 u ) ( + δ2 sinθ u y y sinθ θ θ where Re = Q/ν denotes the Reynolds number. P sinθ +3 θ Re + 1 Re(1 + δy) 2 ], ) + 2δ 3 v θ δ2 u sin 2 θ,

10 Governing equations Scaling Dimensionless equations Boundary conditions Along the air-fluid interface dynamic conditions along the steady free surface y = η(θ) are applied: [ 3δ 2 (η ) 2 + 2(1 + δη) 2 δ(1 + δη)η P P a = δwe [(1 + δη) 2 + δ 2 (η ) 2 ] 3/2 ] δη [ cotθ (1 + δη) 2δ + (1 + δη) 2 + δ 2 (η ) 2 Re[(1 + δη) 2 + δ 2 (η ) 2 (1 + δη) 2 v ] y + δ2 (η ) 2 ( δv + u ) ( ( u (1 + δη)η (1 + δη) θ y + δ δ v ))] (1 + δη) θ u, ( ( u [(1 + δη) 2 δ 2 (η ) 2 ] y + δ δ v )) (1 + δη) θ u ( ( )) v +2δ(1 + δη)η y δ u (1 + δη) θ + δv = 0.

11 Governing equations Scaling Dimensionless equations Boundary conditions Here, We = σh/(ρq 2 ) is the Weber number with σ denoting surface tension, P a the constant ambient air pressure, η = h/h with h being the thickness of the fluid layer, and the prime denotes differentiation with respect to θ. The kinematic condition along the free surface y = η(θ) is given by v = u (1 + δη) η, and the no-slip and impermeability conditions on the surface of the sphere are u = v = 0 at y = 0.

12 Small δ expansion Leading-order problem O(δ) problem For small δ an approximate analytical solution can be constructed by expanding the flow variables in the following series u(y, θ) = u 0 (y, θ) + δu 1 (y, θ) +, v(y, θ) = v 0 (y, θ) + δv 1 (y, θ) +, P(y, θ) = P 0 (y, θ) + δp 1 (y, θ) +, η(θ) = η 0 (θ) + δη 1 (θ) +. Substituting these expansions into the equations of motion and expanding the dynamic conditions in powers of δ leads to a hierarchy of problems.

13 Small δ expansion Leading-order problem O(δ) problem The leading-order problem is governed by the system P 0 y = 3 cosθ Re, subject to 2 u 0 y ( v0 sinθ y + u 0 θ 2 = 3 sinθ, ) + u 0 cosθ = 0, P 0 = P a, u 0 y = 0 at y = η 0, u 0 = v 0 = 0 at y = 0.

14 Small δ expansion Leading-order problem O(δ) problem The solutions are easily found to be P 0 (y, θ) = P a + 3 cosθ Re (η 0 y), u 0 (y, θ) = 3 2 y sinθ(2η 0 y), v 0 (y, θ) = y 2 2 (6η 0 cosθ + 3η 0 sinθ 2y cosθ). Thus, to leading order the pressure is hydrostatic and the velocity in the θ direction, u 0, has a parabolic profile in y which is consistent with flow down an inclined surface at an angle of θ with the horizontal. Further, the velocity component u 0 is symmetric about the plane θ = π/2.

15 Small δ expansion Leading-order problem O(δ) problem The leading-order term for the unknown free surface can be determined by applying the kinematic condition which when transferred from y = η to y = η 0 takes the form v 0 = u 0 η 0 at y = η 0. This leads to the differential equation and solution given by η 0 = 2η 0 cosθ 3 sinθ η 0 (θ) = C sin 2/3 θ. The constant C is found by applying η 0 = h 0 at θ = θ 0 where h 0 and θ 0 are free dimensionless parameters. Thus, C = h 0 sin 2/3 θ 0 and the singularity at θ = 0 is removed. Since we expect the flow to separate from the surface before reaching the bottom of the sphere, the singularity at θ = π is also resolved.

16 Small δ expansion Leading-order problem O(δ) problem The O(δ) problem satisfies the system subject to 2 u 1 y 2 sinθ = Re θ sinθ ( ) P 0 + u2 0 + (Rev 0 2) u 0 2 y, ( v1 y + u 1 θ ( 2y v 0 y + y u 0 θ + 2v 0 ) + u 1 cosθ = ) yu 0 cosθ, u 1 y + η 2 u 0 1 y 2 = u u 0 0 2η 0 y v 0 2η 0 y at y = η 0, u 1 = v 1 = 0 at y = 0.

17 Small δ expansion Leading-order problem O(δ) problem The kinematic condition applied at y = η 0 yields the differential equation for the free surface correction, η 1, given by ( u 0 η 1 + η 0 u 0 y v ) 0 η 1 = v 1 + η 0 η y 0u 0 η 0u 1, subject to η 1 = 0 at θ = θ 0. The solution to the O(δ) problem is significantly more complicated and was obtained using the Maple computer algebra system. For example, the solution for u 1 is:

18 Small δ expansion Leading-order problem O(δ) problem u 1 (y, θ) = y ( Rey 5 sinθ cosθ + 6CRey 4 sin 1/3 θ cosθ 40 ) +15CReη 0y 3 sin 4/3 θ + 60y 2 sinθ + 60y[η 0 cosθ 3C sin 1/3 θ] + F (θ) where F (θ) = 120η 1 sinθ+240c 2 sin 4/3 θ[η 0 cosθ+sinθ] 60C 4 Reη 0 sin 2/3 θ +240C(η 0) 2 sin 1/3 θ 120Cη 0 sin 2/3 θ cosθ 24C 5 Re sin 7/3 θ cosθ. The solution for η 1 can be expressed in the form η 1 (θ) = 512C 2 θ 945 sin 2/3 f (α)dα, θ θ 0 and can be integrated numerically.

19 Small δ expansion Leading-order problem O(δ) problem The function f (α) is defined as: ( f (α) = sin 23/3 α 180C 3 Re sin 2/3 α cos2α 396C 3 Re sin 2/3 α ) +280C cos3α C cosα sin 2/3 α cos3α 2625 sin 2/3 α cosα / (462+cos12α 12 cos10α+66 cos8α 220 cos6α+495 cos4α 792 cos2α).

20 0.55 The leading-order, η 0, and first-order correction, η 1, to the free surface with Re = 1, δ = 0.1, θ 0 = 0.2 and h 0 = η η 0 η 0 + δ η θ

21 1.5 The free surface η = η 0 + δη 1 shown in Cartesian coordinates with Re = 1, δ = 0.1, θ 0 = 0.2 and h 0 = 0.5. y sphere free surface x

22 The dimensionless flow rate per unit width, ˆQ, and dimensionless average streamwise velocity, Û, can be computed using the leading-order solution as follows ˆQ = η0 0 u 0 (y, θ)dy = η 3 0 sinθ, Û = ˆQ η 0 = η 2 0 sinθ. Substituting η 0 = C/ sin 2/3 θ yields ˆQ = C 3 / sinθ and Û = C 2 / sin 1/3 θ. Both the flow rate and average speed decrease as θ increases from θ 0 to π/2 which leads to a decrease in fluid thickness. Note that ˆQ sinθ is constant!

23 The leading-order, u 0, and first-order correction, u 1, to the velocity profile with Re = 1, δ = 0.1, θ 0 = 0.2 and h 0 = 1 at θ = π/4. u u 0 u 0 + δ u y

24 The velocity profile u = u 0 + δu 1 using Re = 1, δ = 0.1, θ 0 = 0.2 and h 0 = 1 at θ = π/4 and θ = π/2. u θ = π/2 θ = π/ y

25 The point of separation can be estimated by using the zero-stress condition: u y = 0 at y = 0. Using u = u 0 leads to 3η 0 (θ) sinθ = 0 and yields the separation angle θ s = π. Including the first-order correction u = u 0 + δu 1 requires solving the equation 3η 0 (θ) sinθ + δ 40 F (θ) = 0, to determine θ s. Here, F (θ) was previously defined. As expected, θ s occurs near the bottom of the sphere with little dependence on Re and δ. When h 0 = 1 and θ 0 = 0.2 separation occurs at θ s while when h 0 = 0.5 and θ 0 = 0.3 separation occurs at θ s

26 The calculation was repeated for the problem of thin flow over a cylinder and very similar streamwise velocity profiles were obtained. However, the free surface was found to vary more rapidly with θ for the sphere than it does for the cylinder. Expressions for η 0 (θ) and η 1 (θ) for the cylinder are given by η 0 (θ) = η 1 (θ) = 7560 sin 1/3 θ D sin 1/3 θ where D = h 0 sin 1/3 θ 0, D 2 θ θ 0 1 sin 11/3 α ( 245 sin 1/3 α cos3α 2205 sin 1/3 α cosα + 70D cos3α D cosα ) +72D 3 Re sin 4/3 α cos2α + 504D 3 Re sin 4/3 α dα.

27 For the cylinder the dimensionless flow rate per unit width, ˆQ, and dimensionless average streamwise velocity, Û, are given by ˆQ = D 3 and Û = ˆQ/η 0 = D 2 sin 1/3 θ. Here, the average streamwise velocity increases as θ increases from θ 0 to π/2 and since the flow rate remains constant the fluid thickness must decrease accordingly.

28 The free surface η = η 0 + δη 1 for the cylinder and sphere with Re = 1, δ = 0.1, θ 0 = 0.2 and h 0 = η cylinder sphere θ

29 Discussed in this work was an analytical investigation of the steady flow of a thin fluid layer over a sphere resulting from a constant discharge from a small hole at the top of the sphere. As an extension the cylindrical case was also solved. The variation in fluid layer thickness is more pronounced for the sphere than it is for the cylinder. This is because the average streamwise velocity increases as the fluid flows over the cylinder and since the flow rate is constant the thickness must therefore decrease. For the sphere both the flow rate and the average streamwise velocity decrease as the fluid flows over the surface which leads to a more rapid decrease in fluid layer thickness. The technique and approach adopted here can be used to model other thin flows that occur in similar settings.

Exercise 5: Exact Solutions to the Navier-Stokes Equations I

Fluid Mechanics, SG4, HT009 September 5, 009 Exercise 5: Exact Solutions to the Navier-Stokes Equations I Example : Plane Couette Flow Consider the flow of a viscous Newtonian fluid between two parallel