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 + r·(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 r·u = 0: Note this is the most general relationship between σ and ru 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·ru = −∇p + µr2u + F @t @u ρ + u·ru = r·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 jρu·ruj ρU2=L UL Re = = = jµr2uj µU=L2 ν If Re 1, then Stokes flow (also called \creeping flow") 2 where ν = µ/ρ is the kinematic viscosity. 0 = −∇p + µr u + F with r·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·ru 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·nj = U: p(x) = µ(U·x)h(r); 2r 3 r=a 2a where r = jxj, 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 = r·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 µr ui = µUi f + 2f =r + 2g + µxi Unxn g + 6g =r We shall need the stress exerted across a spherical surface with r 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. r φ = 0) ur = 2 and uθ = − : r sin θ @θ r sin θ @r u = 2φ − r(x·φ) p = −2µr·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·rr ; 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 r ^ 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 1 Z 1 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 r·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.