Small Particles in a Viscous Fluid

Small particles in a viscous ﬂuid

Course in three parts

Small particles in a viscous ﬂuid 1. A quick course in micro-hydrodynamics

John Hinch 2. Sedimentation of particles

3. Rheology of suspensions May 12, 2014 Good textbook for parts 1 & 2: A Physical Introduction to Suspension Dynamics by Elisabeth Guazzelli, Jeﬀrey F. Morris and Sylvie Pic (Cambridge Texts in Applied Mathematics 2012).

A quick course in micro-hydrodynamics A quick course in micro-hydrodynamics

Stokes equations Stokes equations Continuum mechanics Navier-Stokes equation Simple properties Small Reynolds number Stokes equations Flow past a sphere Simple properties More simple properties Flow past a sphere Greens function More simple properties Eﬀect of small inertia Greens function

Eﬀect of small inertia Continuum mechanics Navier-Stokes equation

Continuum description: mass density ρ(x, t), velocity u(x, t), etc. Newtonian viscous fluids Eulerian, not Lagrangian σij = −pδij + 2µeij , Mass conservation with pressure p(x, t) determined globally by incompressibility ∂ρ + ∇·(ρu) = 0 rather than locally by an equation of state p(ρ) ∂t and strain-rate But for constant density, flows are incompressible 1 ∂ui ∂uj eij = 2 + ∂xj ∂xi ∇·u = 0. Note this is the most general relationship between σ and ∇u which is linear, instantaneous and isotropic. Forces in continuum description: Hence the Navier-Stokes equation (momentum for a Newtonian volume forces F and surface tractions σij nj . viscous fluid) assuming µ constant. ∂u So Cauchy momentum equation ρ + u·∇u = −∇p + µ∇2u + F ∂t ∂u ρ + u·∇u = ∇·σ + F ∂t Boundary conditions u(x) given, or σ·n(x) given

Small Reynolds number Stokes equations

For a ﬂow U over distances L, the Reynolds number is

|ρu·∇u| ρU2/L UL Re = = = |µ∇2u| µU/L2 ν If Re 1, then Stokes ﬂow (also called “creeping ﬂow”) 2 where ν = µ/ρ is the kinematic viscosity. 0 = −∇p + µ∇ u + F with ∇·u = 0. Small Reynolds number, Re 1 if

I small U, e.g. 1 cm/day in oil reservoirs,

I small L, e.g. 10 µm bacteria, Note: Stokes theory for Re 1 usually works for Re < 2. 8 2 I large ν, e.g. 10 m /s molten glass

I e.g. 1 µm water droplet falls under gravity in air at 0.1 mm/s, so Re = 10−5. A quick course in micro-hydrodynamics Simple properties

1. Linear and Instantaneous Stokes equations Dropping nonlinear u·∇u and time-dependent ∂u/∂t leaves Stokes equations linear and instantaneous.

Simple properties E.g. Rigid particle translation at U(t) in unbounded fluid Linear and instantaneous Flow u(x, t) linear& instantanoeus in U(t), also σ(x, t), hence Reversible in time drag force Reversible in space F(t) = A·U(t) with A depending on size, shape, orientation and viscosity. Flow past a sphere 2. Reversible in time More simple properties Apply force F(t) in 0 ≤ t ≤ t1. Now reverse force and its history, i.e. F(t) = −F(2t1 − t) in Greens function t1 ≤ t ≤ 2t1 Then flow u(x, t) and its history reverses. Effect of small inertia Hence all fluid particles return to starting position. Hence cannot swim at Re 1 by reversible flapping.

Simple properties A quick course in micro-hydrodynamics

Stokes equations 3. Reversible in space: Linearity + symmetry of geometry ⇒ certain parts of u vanish. Simple properties E.g. 1. A sphere sedimenting next to a vertical wall does not migrate towards or away from the wall at Re 1. Flow past a sphere The solution E.g. 2. Two equal spheres fall without separating. (Spin?) Method 1 E.g. 3. An ellipsoid (particle with three perpendicular planes of Method 2 symmetry) falls under gravity without rotation in an unbounded Method 3 ﬂow. Method 4 Sedimenting sphere E.g. 4. Two rigid spheres in a shear ﬂow (possibly unequal, Rotating sphere possibly next to a rigid wall) resume their original undisturbed Flow past an ellipsoid streamlines after a collision.

More simple properties

Greens function

Eﬀect of small inertia Flow past a sphere Flow past a sphere

2. Solution method 1 Uniform ﬂow U past a rigid sphere of radius a. The linearity of the Stokes equations means that u(x) must be linear in U. 1. The solution – more important than derivation Further, the problem has spherical symmetry about the centre of the sphere, which take as the origin. 3a a3 3a 3a3 The velocity and pressure ﬁelds must therefore take the forms u = U 1 − − + x(U·x) − + , 4r 4r 3 4r 3 4r 5 u(x) = Uf (r) + x(U·x)g(r), 3aµU·x 3µ p = − and σ·n| = U. p(x) = µ(U·x)h(r), 2r 3 r=a 2a where r = |x|, and f , g and h are functions of scalar r to be Hence the Stokes drag on the sphere is determined. Z 3µ σ·n dS = 4πa2 U = 6πµaU. Now r=a 2a ∂ui 0 0 = Ui xj f /r + δij Unxng + xi Uj g + xi xj Unxng /r. ∂xj

Flow past a sphere Flow past a sphere solution method 1 solution method 1

Contracting i with j, we have the incompressibility condition Hence the general solution of the assumed form linear in u is 0 0 0 = ∇·u = Unxn(f /r + 4g + rg ). 2 −1 1 −3 −3 −5 u(x) = U −2Ar + B + Cr − 3 Dr + x(U·x) A + Cr + Dr , Differentiating again for momentum equation p(x) = µ(U·x) −10A + 2Cr −3 . 2 00 0 00 0 µ∇ ui = µUi f + 2f /r + 2g + µxi Unxn g + 6g /r We shall need the stress exerted across a spherical surface with ∇ p = µU h + µx U x h0/r i i i n n unit normal n = x/r Hence the governing equations give σ·n = U −3Ar + 2Dr −4 + x(U·x) 9Ar −1 − 6Cr −4 − 6Dr −6 f 0/r +4g +rg 0 = 0, f 00+2f 0/r +2g = h and g 00+6g 0/r = h0/r. Eliminating h and then f yields Applying the boundary conditions on the rigid sphere and for the far field, we find the coefficients r 2g 000 + 11rg 00 + 24g 0 = 0. α 3 3 3 Solutions of the form g = r . A = 0, B = 1, C = − 4 a and D = 4 a Substituting, one finds α = 0, −3 and −5, with associated f = −(α + 4)r α+2/(α + 2) and h = −(α + 5)(α + 2)r α. For solution given earlier Flow past a sphere Flow past a sphere

3. Solution method 3 2. Solution method 2 One can show (Papkovich-Neuber) that the general solution of the Use a Stokes streamfunction for the axisymmetric ﬂow Stokes equation can be expressed in terms of a vector harmonic 2 1 ∂Ψ 1 ∂Ψ function φ(x) (i.e. ∇ φ = 0) ur = 2 and uθ = − . r sin θ ∂θ r sin θ ∂r u = 2φ − ∇(x·φ) p = −2µ∇·φ.

The vorticity equation (curl of the momentum equation, to 2 ∂φn ∂ φk eliminate the pressure) is then at low Reynolds numbers σij = 2µ δij − xk . ∂xn ∂xi ∂xj ∂2 sin θ ∂ 1 ∂ D2D2Ψ = 0 where D2 = + . Linearity and spherical symmetry then give ∂r 2 r 2 ∂θ sin θ ∂θ 1 1 φ = AU + BU·∇∇ , 1 2 2 r r The uniform flow at infinity has Ψ = 2 Ur sin θ, so one tries Ψ = F (r) sin2 θ, and finds F = Ar 4 + Br 2 + Cr + D/r. with coefficients A and B to be determined by applying the boundary conditions.

Flow past a sphere Sedimentation of a rigid sphere

Force balance, with densities ρs of sphere and ρf of ﬂuid

4π 3 4π 3 0 = ρs 3 a g − ρf 3 a g − 6πµaU no inertia weight buoyancy Stokes drag 4. Solution method 4

The pressure and vorticity are harmonic functions. So Stokes settling velocity 2∆ρa2g Using linearity and spherical symmetry, they must take the form U = 9µ p = µAU·x/r 3 and ∇ ∧ u = BU ∧ x/r 3. E.g. 1 µm sphere, ∆ρ = 103 kg m−3, water µ = 10−3 Pa s The ﬁnal step to u is tedious. gives U = 2µm/s, i.e. falls through diameter in a second. (Check Re = 10−6)

Drag on a ﬂuid sphere (Student exercise!)

2µf +3µs F = −2π µf aU µf +µs Rotation of a rigid sphere Stokes ﬂow past an ellipsoid

For principle semi-diameters a1, a2, a3 Oberbeck (1876) found 2 2 16πµU1 16πµ(a2 + a3) Force F1 = − 2 and Couple G1 = − 2 2 L + a1K2 3(a2K2 + a3K3) Sphere rotating at angular velocity Ω. where Flow Z ∞ Z ∞ a3 dλ dλ u(x) = Ω ∧ x L = and Ki = 2 r 3 0 ∆(λ) 0 (ai + λ)∆(λ) 2 2 2 2 (A potential ﬂow, so satisﬁes Stokes equations.) with ∆ = (a1 + λ)(a3 + λ)(a3 + λ). Hence couple on sphere – student exercise! For a disk a1 a2 = a3 32 8 F1 ∼ 16πµa2U1, F2 ∼ µa2U2, Gi ∼ µa2Ωi G = 8πµa3Ω 3 3 For a rod a a = a , where ln = ln 2a1 1 2 3 a2 8 3 4πµa1U1 8πµa1U2 16 3 πµa1Ω2 F1 ∼ 1 , F2 ∼ 1 , G1 ∼ 3 πµa1a2Ω1, G2 ∼ 1 ln − 2 ln + 2 ln − 2

Important conclusion Drag ≈ 6πµ with largest diameter.

A quick course in micro-hydrodynamics More simple properties 1. A useful result

Stokes equations Let uS (x) be a Stokes ﬂow with F = 0 in V , and let u(x) be any other incompressible ﬂow, then Z Z Simple properties S S 2µeij eij dV = σij nj ui dA. V S Flow past a sphere Because More simple properties S S S S S 2µeij = σij + p δij and p δij eij = p ∇·u = 0 (1) A useful result so 2µeS e = σS e . (2) Minimum dissipation ij ij ij ij Uniqueness And Geometric bounding σS = σS so (3) Reciprocal theorem ij ji S Symmetry of resistance matrix S S ∂ui ∂ S ∂σij σij eij = σij = σ ijui − ui (4) Faxen’s formula ∂xj ∂xj ∂xj =F=0 Greens function Hence result by divergence theorem.

Eﬀect of small inertia More simple properties More simple properties 2. Minimum dissipation 2. Minimum dissipation

Let u(x) and uS (x) be two incompressible flows in V , both satisfying the same boundary condition u = uS = U(x) give on S. S Let u also satisfy the Stokes equation with F = 0 in V . Hence Z Z S S Then 2µeij eij dV ≥ 2µeij eij dV , V V Z Z S S S 2µeij eij dV = 2µeij eij dV i.e. the Stokes flow u (x) has the minimum dissipation out of all V V Z Z incompressible flows satisfying the boundary condition S S S S + 2µ(eij − eij )(eij − eij ) dV + 4µeij (eij − eij ) dV . V V Hence e.g. drag larger at non-zero Reynolds number. ←−positive−→

The last integral is of the form of the useful result Warning: Same geometry. Cannot select geometry by minimum dissipation. Z Z S S S S 4µeij (eij − eij ) dV = 2σij nj (ui − ui ) dA = 0 by bc V S

More simple properties More simple properties 3. Uniqueness 4. Geometric bounding

If u1(x) and u2(x) are two Stokes flows in V satisfying the same Rigid cube, sides of length 2L, moving at U, drag F. boundary conditions, then minimum dissipation gives Let uS (x) be Stokes flow outside cube, V . Then dissipation Z 1 2 1 2 2µ(eij − eij )(eij − eij ) dV = 0 Z V S S 2µeij eij dV = rate of working by surface forces = −U·F. V Hence e1 − e2 = 0 in V , √ ij ij Cube just contained by sphere radius a = 3L, also moving at U. i.e. u1 − u2 is strainless, i.e. a solid body translation + rotation, i.e. zero by the boundary conditions. Hence Define second flow ( 1 2 the Stokes flow for sphere outside sphere, u (x) = u (x) in V . u(x) = U in gap. Hence Stokes flows are unique. More simple properties More simple properties 4. Geometric bounding 5. Reciprocal theorem

For this second flow For the same volume V , let u1 be the Stokes flow with volume Z Z forces f1 satisfying boundary conditions u1 = U1. 2µeij eij dV = 2µeij eij dV because e = 0 in gap, Let u be Stokes flow in V with f and U . V r>a √ 2 2 2 = rate of working by sphere = 6πµ 3LU·U Then by the useful result Hence minimum dissipation bounds drag F on cube Z Z Z √ u1 ·f2 dV + U1 ·σ2 ·n dA = 2µe1 : e2 dV −F·U ≤ 6πµ 3LU·U V S V Z Z = u2 ·f1 dV + U2 ·σ1 ·n dA Similarly for sphere just contained inside cube V S

6πµLU·U ≤ −F·U i.e. work done by one velocity on the forces of the other is vice versa. Greens theorem in any other subject Student excercise: bound for tetrahedron (not so tight).

More simple properties More simple properties 6. Reciprocal theorem – application to resistance matrix 6. Reciprocal theorem – application to resistance matrix

General rigid body motion in ﬂuid at rest a inﬁnity, translating at U(t) and rotating (about a selected point) Ω(t). B = CT means By linearity and instantaneity, the force F(t) and couple G(t) (about the same selected point) force due to rotation = couple due to translating.

F AB U T T = A = A & D = D for symmetric cube means G CD Ω A & D diagonal, and B = CT = 0 The Reciprocal theorem gives for the two rigid body motions Thus drag on a cube is parallel to velocity, U1 ·F2 + Ω1 · G2 = U2 ·F1 + Ω2 · G1 also for symmetric tetrahedron. True all U etc, so 1 Need “corkscrew” feature for B 6= 0. A = AT , B = CT and D = DT . More simple properties More simple properties 7. Reciprocal theorem – application to Faxen’s formula 7. Reciprocal theorem – application to Faxen’s formula

∞ A force-free sphere placed in arbitrary flow u (x) moves with what Z Z ∞ + ∞ velocity V? Forces which generate u kept constant. RHS = U2 · σ ·n dA − σ ·n dA = 0 − 0 = 0, Let u+ be Stokes flow with sphere inserted in u∞. + ∞ as both integrals are force on sphere. Then disturbance flow is u1 = u − u . + ∞ Hence Now same forces for u & u , so f1 = 0. Also u1 → 0 far from Z Z sphere. 3µ + ∞ LHS = − U2 · u dA − u dA = RHS = 0 Let u2(x) be flow outside a sphere translating at U2 with f2 = 0. a =V Applying Reciprocal theorem ∞ Z Z Z Z For all U2, so velocity of sphere inserted into u (x) is u1· f2 dV + u1· σ2 ·n dA = u2· f1 dV + u2 ·σ1·n dA. =0 =0 =U Z =−3µu2/a 2 1 ∞ V = 2 u (x) dA + ∞ 4πa r=a Now look at RHS and then LHS, using u1 = u − u .

More simple properties A quick course in micro-hydrodynamics 7. Reciprocal theorem – application to Faxen’s formula

Stokes equations Z 1 ∞ V = 2 u (x) dA Simple properties 4πa r=a Flow past a sphere Finally use a Taylor series

∞ ∞ ∞ 1 ∞ u (x) = u (0) + x·∇u |0 + 2 xx : ∇∇u |0 + ··· More simple properties

Greens function Integrating over the sphere, the odd terms vanish by symmetry, so Stokeslet ∞ a2 2 ∞ Integral representation V = u (0) + ∇ u |0 6 Slender-body theory with higher even terms vanishing by ∇2n(Stokes equations) = 0. Eﬀect of small inertia Greens function for Stokes equations Greens function for Stokes equations or ‘Stokeslet’ Far ﬁeld for a sphere

For a point momentum source Far from the sphere r a, the ﬂow is ∇·u = 0 0 = −∇p + µ∇2u + Fδ(x) 3a a3 3a 3a3 u = U 1 − − + x(U·x) − + , 4r 4r 3 4r 3 4r 5

Solution – more important than derivation But the drag is F = −6πµaU, i.e.

1 1 1 1 1 1 u(x) = F·G(x) = F + (F·x)x u − U ∼ F + (F·x)x 8πµ r r 3 8πµ r r 3 3 1 σ(x) = F·K(x) = − F·xxx 5 4π r Hence far-ﬁeld due to force is universal, independent of particle G is called the ‘Oseen tensor’. shape.

Already seen this in ﬂow past a sphere:

Greens function for Stokes equations Greens function for Stokes equations Integral representation Integral representation – Boundary integral Method

To solve

0 ∇·u = 0 Letting x in V tend onto the surface S yields an integral equation for the unknown u (or σ·n) on S in terms of the known σ·n (or u) 0 = −∇p + µ∇2u + f(x) on S. 0 R 1 0 1 0 with boundary conditions on u(x) or σ(x)·n. Delicate limit x → S: K·n → + 2 u for x in V , − 2 u for x outside V , more complex at corners on S. Use the Reciprocal theorem (Greens theorem) with u1 for the 0 unknown flow and u2 for the Greens function for point source at x Basis of numerical Boundary Integral Method. Z Advantage: fewer points on surface than in volume, and no infinity. u(x0) = G(x − x0)·f dV V Disadvantages: Special attention needed in numerical evaluation of forces in V Z singular integrals, and there are often eigensolutions, e.g. constant + G(x − x0)·σ(x)·n − (K(x − x0)·n)·u dA pressure induces no flow. S forces on S dipoles on S Greens function for Stokes equations Greens function for Stokes equations Integral representation - for a suspended drop Slender-body theory

Extension to a drop of viscosity λµ surrounded by a ﬂuid of For slender bodies, approximate the distribution of forces σ·n on viscosity µ. the surface S by a distribution of forces f(s) along the centreline One knows the jump across the interface of the normal viscous X(s) in −L ≤ s ≤ L stress Z L σ·n = γκn u(x0) = u∞(x0) + G(X(s) − x0)·f(s) ds −L with surface tension γ and surface curvature n. Satisfy the boundary condition by evaluating at distance R(s0) Adding λ times the integral equation for the interior to that for the from centreline at s = s0, either numerically exterior, one reduce the stress contribution to its diﬀerence or asymptotically for 1 0 ∞ 0 2 (1 + λ)u(x ) = u (x ) Z Z 2πµ 0 0 ∞ 0 0 f(s0) ∼ 2I − X X ·(U(s0) − u (X(s0))) − G(x − x )·γκn dA − (1 − λ) K(x − x )·n·u dA ln L S S R

Effects of small inertia on flow past a sphere Effects of small inertia on flow past a sphere Whitehead paradox Oseen equation

Look more carefully at Stokes solution. In far field Flow U past a sphere of radius a. At Re 1, first approximation is given by Stokes flow. µUa ρU2a Viscous terms = O , while interial terms = O( Attempt to find correction as regular perturbation fails. r 3 r 2 In far field so at r = µ/U inertial terms no longer small. Ua ρU2a u = U + O , so ρu·∇u = O r r 2 Fortunately in far field u = U + u0 with u0 small, so can linearise Navier-Stokes A correction u2 forced by this ρU·∇u0 = −∇p0 + µ∇2u0 + Fδ(x) 2 −∇p2 + µ∇ u2 = ρu·∇u|Stokes where in far field sphere appears a point force. would give 2 u2 = O(ρU a/µ) Solve by Fourier transforms or representation which does not decay. u0 = ∇φ + ν∇χ − Uχ and p0 = −ρU·∇φ. Effects of small inertia on flow past a sphere Effects of small inertia on flow past a sphere Oseen equation solved Oseen wake

Look far from sphere r ν/U near downstream axis Find 2 2 0 3a − U(x +y ) u ∼ −U e 4νz 3aν 3a U·x − Ur φ = − (point volume source) and χ = e( 2ν 2ν ) 2z 2r 2r i.e. a wake diffusing to r = pν(t = z/U) with mass flux deficit

Look nearer to sphere a r ν/U, r −2 terms cancel and Z ρu0 dxdy = −6πµUa =momentum deficit /U 1 3Ua u0 ∼ Stokeslet + U uniform flow r 4ν Helpful idea at higher Re 3 Impacts on time-dependent flows – Basset history wrong Hence drag increases by 1 + 8 Re. Missing mass flux in wake goes to point source φ-flow.

A quick course in micro-hydrodynamics

Stokes equations

Simple properties

Flow past a sphere

More simple properties

Greens function

Eﬀect of small inertia