3D DLM/FD METHODS FOR SIMULATING THE MOTION OF SPHERES IN BOUNDED SHEAR FLOWS OF OLDROYD-B FLUIDS

Transcription

1 3D DLM/FD METHODS FOR SIMULATING THE MOTION OF SPHERES IN BOUNDED SHEAR FLOWS OF OLDROYD-B FLUIDS A Dissertation Presented to the Faculty of the Department of Mathematics University of Houston In Partial Fulfillment of the Requirements for the Degree Doctor of Philosophy By Shang-Huan Chiu April 217

2 3D DLM/FD METHODS FOR SIMULATING THE MOTION OF SPHERES IN BOUNDED SHEAR FLOWS OF OLDROYD-B FLUIDS Shang-Huan Chiu APPROVED: Dr. Tsorng-Whay Pan, Chairman Department of Mathematics, University of Houston Dr. Roland Glowinski, Department of Mathematics, University of Houston Dr. Jiwen He, Department of Mathematics, University of Houston Dr. Dong Liu, Department of Mechanical Engineering, University of Houston Dean, College of Natural Sciences and Mathematics ii

3 3D DLM/FD METHODS FOR SIMULATING THE MOTION OF SPHERES IN BOUNDED SHEAR FLOWS OF OLDROYD-B FLUIDS An Abstract of a Dissertation Presented to the Faculty of the Department of Mathematics University of Houston In Partial Fulfillment of the Requirements for the Degree Doctor of Philosophy By Shang-Huan Chiu April 217 iii

4 Abstract iv

5 Contents 1 Introduction 1 2 3D DLM/FD methods for simulating the motion of spheres in bounded shear flows of Newtonian fluids Stokes Equations and Newtonian Model DLM/FD Method for Simulating Fluid-particle Interaction in Stokes Flow The Governing Equations Variational Formulation Finite Element approximation and Operator Splitting scheme Finite Element approximation Collocation Boundary Method and Immersed Boundary Method Operator Splitting scheme On the solution of the subproblems from operator splitting Solution of the rigid body motion enforcement problems Numerical results v

6 CONTENTS Rotation of a single particle Sedimentation of a single particle Two balls interaction in an one wall-driven bounded shear flow Two balls interaction in a two wall-driven bounded shear flow D DLM/FD methods for simulating the motion of spheres in bounded shear flows of Oldroyd-B fluids Several Models of Non-Newtonian Fluid Stokes Equations Newtonian Model The UCM-model The Oldroyd-B model The Carreau model DLM/FD method for simulating fluid-particle interaction in stokes flow The governing equations DLM/FD formulation Numerical methods Finite element approximation Operator splitting scheme Logarithm of conformation tensor Operator splitting scheme of Carreau model On the solution of the subproblems from operator splitting Solution of the advection subproblems Solution of the rigid body motion enforcement problems Numerical results Rotation of a single particle vi

7 CONTENTS Migration of a single particle in an one wall driven bounded shear flow Two ball interacting with large initial distance in a two wall driven bounded shear flow Two ball interacting with small initial distance in an one wall driven bounded shear flow Concluions and Future work Concluions Future work Bibliography 77 vii

8 List of Figures The tumbling motion as W i =.5: the initial distance between two particles is 2r+gap where r =.1, gap=3h.for each frame, the horizontal direction is X 1 axis and vertical direction is X 3 axis The kayaking motion as W i = 1.: the initial distance between two particles is 2r+gap where r =.1, gap=3h. For each frame, the horizontal direction is X 2 axis and vertical direction is X 3 axis Particles alignment in polyisobutylene solution(j. Michele et al. Rheol. Acta 1977) viii

9 List of Tables ix

10 CHAPTER 1 Introduction Suspensions of particles in fluids appear in many applications of chemical, biological, petroleum, and environmental areas. For the dynamics of rigid non-brownian particles suspended in viscoelastic fluids, peculiar phenomena of the particle motion and pattern induced by fluid elasticity have been reviewed on theoretical predictions, experimental observations and numerical simulations from the existing papers in [6] In Newtonian fluid, plenty numerical and experimental results have been published. Several topics are considered, such as the random displacements resulting 1

11 from particle encounters under creeping-flow conditions and the rotation of a neutrally buoyant particle in simple shear flow. The displacements from particle encounters lead to hydrodynamically induced particle migration, which constitutes an important mechanism for particle redistribution in the suspending fluid (see, e.g., [21] and the references therein). Binary encounters of particles is a phenomenon have been found that two balls either pass each other or swapping their cross streamline position. But particle suspensions in viscoelastic fluids have different behaviors, e.g., strings of spherical particles aligned in the flow direction (e.g., see [12, 16, 19, 14]) and 2D crystalline patches of particles along the flow direction [15] in shear flow. As mentioned in [17], these flow-induced self-assembly phenomena have great potency for creating ordered macroscopic structures by exploiting the complex rheological properties of the suspending fluid as driving forces, such as its shear-thinning and elasticity. Furthermore, shear-thinning of the viscosity seems to be the key rheological parameter that determines the overall nature of the hydrodynamical interactions, rather than the relative magnitude of the normal stress differences. Same conclusion about the role of shear-thinning on the aggregation of many particles has been reported in [16, 19]. There are numerical studies of the two particle interaction and aggregation in viscoelastic fluids(e.g., see [8, 3, 2]). Several non-newtonian fluid models in bounded shear flow have been considered, such as Oldroyd-B fluid, Giesekus fluid. For the encounter of two balls in a bounded shear flow, the trajectories of the two ball mass centers are consistent with those obtained in [2]. We have further 2

12 tested the cases of two balls for the Weissenberg number up to 1 and obtained they either pass, return, or tumble in a bounded shear flow with two moving wall. The trajectories of the two ball mass centers lose the symmetry due to the effect of elastic force arising from viscoelastic fluids for the higher values of the Weissenberg number. For the interaction of the two balls in one wall driven shear flow, two balls form a loosely connected chain if the initial gap between two balls is small enough and then these two balls keep rotating with respect to the midpoint between their mass centers and migrate toward the moving wall. DLM/FD Method To simulate the interaction of neutrally buoyant balls in 3D bounded shear flow of Oldroyd-B fluids, we have generalized a distributed Lagrange multiplier/fictitious domain method (DLM/FD) developed in [13] for simulating the motion of neutrally buoyant particles in Stokes flows of Newtonian fluids from 2D to 3D and then combined such method with the operator splitting scheme and matrix-factorization approach for treating numerically the constitutive equations of the conformation tensor of Oldroyd-B fluids. In this matrix-factorization approach, which is the technique close to the one developed by Lozinski and Owens in [11], 3

13 CHAPTER 2 3D DLM/FD methods for simulating the motion of spheres in bounded shear flows of Newtonian fluids 2.1 Stokes Equations and Newtonian Model Consider the Stokes equations in (, T ) describing the 3-dimensional motion of an incompressible and isothermal fluid without body force: σ = ρ f g, u =. 4

14 2.2. DLM/FD METHOD FOR SIMULATING FLUID-PARTICLE INTERACTION IN STOKES FLOW where σ = pi + τ with presure p and stress tensor τ, u = (u 1, u 2, u 3 ) is the flow velocity, g is the gravity, and ρ f is the density of fluid. The constitutive equation of the stress tensor is given by τ = 2µD(u) where µ is the Newtonian viscosity of the fluid and D(u) is the symmetric rate of strain tensor D(u) = u + ( u)t, 2 with the Jacobian of the velocity u, u 1 x u u = 2 x u 3 x u 1 y u 2 y u 3 y u 1 z u 2 z u 3 z 2.2 DLM/FD Method for Simulating Fluid-particle Interaction in Stokes Flow The Governing Equations Let be a bounded domain in R 3 and let Γ be the boundary of. We suppose that is filled with a viscous fluid with density ρ f and contains N moving particles of density ρ s. Let B(t) = N i=1 B i (t) where B i (t) is the i-th soild particle in the fluid for 5

15 2.2. DLM/FD METHOD FOR SIMULATING FLUID-PARTICLE INTERACTION IN STOKES FLOW i = 1, 2,, N. We denote by γ i (t) the boundary B i (t) of B i (t) for i = 1, 2,, N and let γ(t) = N i=1 γ i (t). For some T >, the governing equations for the fluid-particles system is as follows: For the motion of fluid, we consider the Stokes equations in Oldroyd-B model: σ = ρ f g in \ B(t), t (, T ), (2.2.1) u = g u = in \ B(t), t (, T ), (2.2.2) on Γ (, T ), with Γ g ndγ =, (2.2.3) u(x, t) = V i (t) + ω i (t) G i (t)x, x γ i (t), i = 1, 2,, N, (2.2.4) where u is the flow velocity, p is the presure, g is the gravity, σ = pi + τ s, τ s = 2µ f D(u) is a Newtonian stress tensor, ρ f is the density of fluid and µ f is the solvent viscosity of the fluid, and n is the unit normal vector pointing outward to the flow region. In (2.2.4), we assume a no-slip condition on γ(t): u(x, t) = V i (t) + ω i (t) G i (t)x, x γ i (t), i = 1, 2,, N, where V i is the translation velocity, ω i is the angular velocity, G i is the center of mass and x is a point on the surface of the particle. 6

16 2.2. DLM/FD METHOD FOR SIMULATING FLUID-PARTICLE INTERACTION IN STOKES FLOW In (2.2.1), we use σ = pi + 2µ f D(u) and obtain p 2µ f D(u) = ρ f g. (2.2.5) The motion of particle satisfies the Euler-Newton s equations: v i (x, t) = V i (t) + ω i (t) G i (t)x, {x, t} B i (t) (, T ), i = 1, 2,, N, (2.2.6) dg i dt = V i, (2.2.7) M i dv i dt d(i i ω i ) dt = M i g + F i + F r i, (2.2.8) = T i + G i x r F r i, (2.2.9) G i () = G i, V i () = V i, ω i () = ω i, (2.2.1) for i = 1, 2,, N, where M i and I i are the mass and the moment of inertia of the i-th particle, respectively; F i and T i are the hydrodynamical force and torque imposed on the i-th particle by the fluid, and F r i is a short range repulsion force imposed on the i-th particle by other particles and wall to prevent particle/particle and particle/wall penetration. The hydrodynamical force F i and torque T i imposed on the i-th particle by the fluid are given by F i = σn dγ, γ T i = G i x σn dγ. γ 7

17 2.3. VARIATIONAL FORMULATION G i x r F r i is a torque acting on the point x r where F r i applies on the i-th particle. 2.3 Variational Formulation For convenience of derivation, we assume there is only one particle in the fluid, that is, we set B(t) the soild particle in the fluid, γ(t) the boundary B(t) of B(t), G(t) the center of mass of this particle. In the equations of the motion of particle, we set V the translation velocity of the particle B(t), ω the angular velocity of the particle B(t), M p and I p the mass and the moment of inertia of the particle B(t), respectively; F and T the hydrodynamical force and torque imposed on the particle B(t) by the fluid, respectively, and F r a short range repulsion force imposed on the particle B(t). To obtain a variational formulation for above problem (2.2.5),(2.2.2) (2.2.4), we define the following function spaces { W g (t) = v v (H 1 ( \ B(t))) 3, v = g (t) on Γ, v = V(t) + ω(t) G(t)x } on B(t), { W (t) = (v, Y, θ) v (H 1 ( \ B(t))) 3, v = on Γ, v = Y + θ G(t)x on B(t), with Y R 3 and { L 2 ( \ B(t)) = q q L2 ( \ B(t)), \B(t) } q dx =. Applying the virtual power principle to the system (2.2.5),(2.2.2) (2.2.4) obtains the following variational formulation: 8

18 2.3. VARIATIONAL FORMULATION For a.e. t >, find u(t) W g (t), p(t) L 2 ( \ B(t)),V(t) R 3, G(t) R 3, ω(t) R 3, such that p v dx + 2µ f D(u) : D(v) dx \B(t) \B(t) ( ) ( dv + M p dt M pg F r dω Y + I p dt ) Gx r F r θ = ρ f g v dx, (v, Y, θ) W (t), \B(t) (2.3.11) \B(t) q u dx =, q L 2 ( \ B(t)), (2.3.12) dg dt = V, (2.3.13) u(x, ) = u (x), x \ B(), (2.3.14) G(x, ) = G (x), V(x, ) = V (x), ω(x, ) = ω (x), (2.3.15) To obtain an equivalent fictitious domain formulation, first we fill the particle B(t) with a fluid of density ρ f and suppose that this fluid follows the same rigid body motion as B(t) itself, which is u(x, t) = V(t) + ω(t) G(t)x, x B(t). (2.3.16) Define a function space { W (t) = (v, Y, θ) (v \B(t), Y, θ) W (t), v(x, t) = Y + θ G(t)x } x B(t). 9

19 2.3. VARIATIONAL FORMULATION Suppose particle B is made of an homogeneous material of density ρ f which follows ρ f B(t) g v dx = ρ f ρ s M p g Y, (v, Y, θ) W (t), (2.3.17) v = in B(t), (v, Y, θ) W (t), (2.3.18) u = in B(t) and D(u) = in B(t). (2.3.19) To obtain a fictitious domain formulation, we define the following function spaces V g (t) = { v v (H 1 ()) 3, v = g (t) on Γ }, { L 2 () = q q L2 (), } q dx =. Combining (2.3.11)-(2.3.15) with (2.3.16)-(2.3.19), we obtain the fictitious domain formulation as follows: For a.e. t >, find find u(t) V g (t), p(t) L 2 (), V(t) R 3, G(t) R 3, ω(t) R 3, such that p v dx + 2µ f D(u) : D(v) dx dv +M p dt Y + I dω p dt θ Fr Y Gx r F r θ ( = ρ f g v dx + 1 ρ ) f M p g Y, ρ s (v, Y, θ) W (t), (2.3.2) q u dx =, q L 2 (), (2.3.21) 1

20 2.3. VARIATIONAL FORMULATION dg dt = V, (2.3.22) u(x, t) = V(t) + ω(t) G(t)x, x B(t). (2.3.23) u(x, ) = u (x), x \ B(), (2.3.24) G(x, ) = G (x), V(x, ) = V (x), ω(x, ) = ω (x), (2.3.25) To relax the rigid body motion condition (??), we introduce a Lagrange multiplier, λ Λ(t) = (H 1 (B(t))) 3, and for any µ (H 1 (B(t))) 3 and v (H 1 ()) 3 : µ, v Λ(t) = (µ v + µ v) dx. B(t) We obtain the fictitious domain formulation with Lagrange multiplier as follows: For a.e. t >, find find u(t) V g (t), p(t) L 2 (), V(t) R 3, G(t) R 3, ω(t) R 3, λ Λ(t) such that dv p v dx + 2µ f D(u) : D(v) dx + M p dt Y + I dω p dt θ F r Y Gx r F r θ λ, v Y θ Gx Λ(t) ( = ρ f g v dx + 1 ρ ) f M p g Y, ρ s (v, Y, θ) (H 1 ()) 3 R 3 R 3, (2.3.26) q u dx =, q L 2 (), (2.3.27) dg dt = V, (2.3.28) 11

21 2.4. FINITE ELEMENT APPROXIMATION AND OPERATOR SPLITTING SCHEME u(x, ) = u (x), x \ B(), (2.3.29) µ, u(t) V(t) ω(t) G(t)x =, µ Λ(t), (2.3.3) Λ(t) G(x, ) = G (x), V(x, ) = V (x), ω(x, ) = ω (x). (2.3.31) Since u is divergence free and satisfies the Dirichlet boundary conditions on Γ, we obtain 2 D(u) : D(v) dx = u : v dx, v (H()) Finite Element approximation and Operator Splitting scheme Finite Element approximation For the purpose of finding an approximation solution of problem (2.3.26)-(2.3.31), we need a partition of the flow region R 3. We use an uniform finite element mesh for and let h be the uniform finite element mesh size for the velocity field, T h be a tetrahedrization of, and T 2h be twice coarser than T h. For space discretization, we have chosen P 1 -iso-p 2 finite elements for the velocity field and P 1 finite elements for the pressure where P i is the space of the polynomials in three variables of degree i for i = 1, 2. 12

22 2.4. FINITE ELEMENT APPROXIMATION AND OPERATOR SPLITTING SCHEME Then we define the following function spaces: V h = {v h vh ( C ( ) ) 3, vh T (P 1 ) 3, T T h }, V gh (t) = {v h v h V h, v h Γ = g h (t) }, V h = {v h v h V h, v h Γ = }, L 2 h = { q h qh C ( ), q h T P 1, T T 2h }, and L 2 h = { } q h q h L 2 h, q h dx =, where g h (t) is an approximation of g h (t) satisfying g h (t) ndγ =, Γ and Γ h = {x x Γ, g h(x, t) n(x) < }. For the space of Lagrange multipier Λ(t), we need to define a finite dimensional space to approach Λ(t). Let {ξ i } N i=1 be a set of points from B(t) which cover B(t) evenly. We define the discrete Lagrange multiplier space by Λ h (t) = { µ h µ h = } N µ i δ(x ξ i ), µ i R 3, i = 1,, N, i=1 where x δ(x ξ i ) is the Dirac measure at x = ξ i. For different approaches, 13

23 2.4. FINITE ELEMENT APPROXIMATION AND OPERATOR SPLITTING SCHEME there are two different definitions of discretize scalar pairung <, > Λh (t). We will introduce these two scalar pairings in next section. Baesd on the finite dimensional spaces above, we obtain the following approximation of problem (2.3.26)- (2.3.31) : For a.e. t >, find find u h (t) V gh (t), p(t) L 2 h, V(t) R3, G(t) R 3, ω(t) R 3, λ h Λ h (t) such that p v dx + 2µ f u h : v dx dv +M p dt Y + I dω p dt θ Fr Y Gx r F r θ ( = 1 ρ ) f M p g Y + λ h, v Y θ Gx ρ s (v, Y, θ) V h R 3 R 3, Λ h (t), (2.4.32) q u h dx =, q L 2 h, (2.4.33) dg dt = V, (2.4.34) u h (x, ) = u h (x), x \ B h (), (2.4.35) µ h, u h (t) V(t) ω(t) G(t)x =, µ h Λ h (t), (2.4.36) Λ h (t) G(x, ) = G (x), V(x, ) = V (x), ω(x, ) = ω (x), (2.4.37) 14

24 2.4. FINITE ELEMENT APPROXIMATION AND OPERATOR SPLITTING SCHEME where u h is an approximation of u such that q u h dx =, q L 2 h Collocation Boundary Method and Immersed Boundary Method When we deal with the boundary of the particle B(t), there are several methods have been considered for discretization. Here we introduce two of them, which are collocation boundary method and immersed boundary method. For collocation boundary method, we define the scalar pairing <, > Λh (t) as follows: N µ h, v h Λh (t) = µ i v h (x i ), µ h Λ h (t), v h V gh (t) or V h, i=1 By using the above scalar pairing <, > Bh (t), the rigid body motion of particle B(t) is forced via a collocation boundary method. For immersed boundary method, we define the scalar pairing <, > Λh (t) as follows: N µ h, v h Λh (t) = M µ i v h (x j )D h (x j ξ i )h 3, µ h Λ h (t), v h V gh (t) or V h, i=1 j=1 where {x j } M j=1 are the grid points of the finite elements for the velocity, and the 15

25 2.4. FINITE ELEMENT APPROXIMATION AND OPERATOR SPLITTING SCHEME function D h (X ξ i ) is defined as D h (X ξ i ) = δ h (X 1 ξ i1 )δ h (X 2 ξ i2 )δ h (X 3 ξ i3 ) with X = (X 1, X 2, X 3 ) and ξ i = (ξ i1, ξ i2, ξ i3 ), and the one-dimensional discrete δ h defined by δ h (z) = z 8h h z h 1 8h z h z h 4 ( ) 2 z, z h, h ( ) 2 z, h z 2h, h, otherwise. By using the above scalar pairing <, > Λh (t), the rigid body motion of particle B(t) is forced via an immersed boundary method Operator Splitting scheme Applying the Lie s scheme to the discrete analogue of the problem (2.3.26)-(2.4.37) and backward Euler s method, we obtain: Given u = u h, G = G, V = V, ω = ω. For n, u n, G n, V n, ω n are known, we predict the position and the translation velocity of the center of mass as follows. dg dt = V(t), (2.4.38) 16

26 2.4. FINITE ELEMENT APPROXIMATION AND OPERATOR SPLITTING SCHEME dv M p =, dt (2.4.39) dω I p =, dt (2.4.4) V(t n ) = V n, ω(t n ) = ω n, G(t n ) = G n, (2.4.41) for t n < t < t n+1. Then set V n+ 1 2 = V(t n+1 ), ω n+ 1 2 = ω(t n+1 ), and G n+ 1 2 = G(t n+1 ). and we get B n+ 1 2 h based on the center of particle G n+ 1 2 and take ω n+ 1 2 = ω n. Then we enforce the rigid body motion in B n+ 1 2 h and solve u n+1 and p n+1 simultaneously as follows: Find u n+1 Vg n+1 h, p n+1 L 2 h, λn+1 Λ n+1, V n+1 R 3, ω n+1 R 3 such that h p n+1 v dx + 2µ f u n+1 : v dx V n+1 V n+ 1 2 ω n+1 ω n M p Y + I p θ ( t t = 1 ρ ) f M p g Y + λ n+1, v Y θ G n+ 1 2 x ρ s (v, Y, θ) V h R 3 R 3, Λ n+1 h, (2.4.42) q u n+1 dx =, q L 2 h, (2.4.43) µ, u n+1 V n+1 ω n+1 G n+ 1 2 x =, µ Λ n+ 1 2 h. (2.4.44) Λ n+1 h Finally we set G n+1 = G n In the above, Vg n+1 h = V gh (t n+1 ), Λ n+1 h = Λ h (t n+1 ), and B n+s h = B h (t n+s ). 17

27 2.5. ON THE SOLUTION OF THE SUBPROBLEMS FROM OPERATOR SPLITTING 2.5 On the solution of the subproblems from operator splitting Solution of the rigid body motion enforcement problems In system (2.4.42)-(2.4.44), there are two multipliers. p and λ. We have solved this system via an Uzawa-conjugate gradient method driven by both multipliers. The general problem is as follows: Find u V gh, p L 2 h, λ Λ h, V R 3, ω R 3 such that = V V p v dx + µ f u : v dx + M p t ( 1 ρ ) f M p g Y + λ, v Y θ Gx ρ s (v, Y, θ) V h R 3 R 3, Λ h, Y + I p ω ω t θ (2.5.45) q u dx =, q L 2 h, (2.5.46) µ, u V ω Gx =, µ Λ h. (2.5.47) Λ h Applying the following Uzawa-conjugate gradient algorithm operating in the space L 2 h Λ h to solve the sysyem (2.5.45)-(2.5.47): Assume p L 2 h and λ Λ h are given. We solve the problem: 18

28 2.5. ON THE SOLUTION OF THE SUBPROBLEMS FROM OPERATOR SPLITTING Find u V gh, V R 3, ω R 3 satisfying µ f u : v dx = p v dx + λ, v, Λ h v V h ; u V gh, (2.5.48) M p V V t and then compute Y = I p ω ω t ( 1 ρ ) f M p g Y λ, Y, Y R 3, (2.5.49) ρ Λ s h θ = λ, θ Gx Λ h, θ R 3, (2.5.5) g 1 = u ; (2.5.51) next find g 2 Λ h satisfying µ, g 2 Λ h = µ, u V ω Gx, µ Λ h, (2.5.52) Λ h and set w 1 = g 1, w 2 = g 2. (2.5.53) Then for k, assuming that p k, λ k, u k, V k, ω k, g k 1, g k 2, w k 1 and w k 2 are known, compute p k+1, λ k+1, u k+1, V k+1, ω k+1, g k+1 1, g k+1 2, w 1 k+1 and w k+1 2 as follows: µ f u k : v dx = w1 k v dx + w2, k v, Λ h v V h ; u k V gh, (2.5.54) 19

29 2.5. ON THE SOLUTION OF THE SUBPROBLEMS FROM OPERATOR SPLITTING and then compute M p V k t Y = w k 2, Y Λ h, Y R 3, (2.5.55) ω k I p t θ = w2, k θ Gx, θ R 3, (2.5.56) Λ h g k 1 = u k ; (2.5.57) next find g k 2 Λ h satisfying µ, g k 2 Λ h = µ, u k V k ω k Gx, µ Λ h, (2.5.58) Λ h and compute and ρ k = g k 1 2 dx + g k 2, g2 k Λ h g k 1w1dx k +, (2.5.59) g k 2, w2 k Λ h p k+1 = p k ρ k w 1 k, (2.5.6) λ k+1 = λ k ρ k w k 2, (2.5.61) u k+1 = u k ρ k u k, (2.5.62) V k+1 = V k ρ k V k, (2.5.63) ω k+1 = ω k ρ k ω k, (2.5.64) g k+1 1 = g k 1 ρ k g k 1, (2.5.65) g k+1 2 = g k 2 ρ k g k 2. (2.5.66) 2

30 2.6. NUMERICAL RESULTS If g k dx + g2 k+1, g2 k+1 g 1 2 dx + g 2, g2 Λ h Λ h ε, (2.5.67) then take p = p k+1, λ = λ k+1, u = u k+1, V = V k+1, and ω = ω k+1. Otherwise, compute and set γ k = g k dx + g2 k+1, g2 k+1 g k 1 2 dx + g k 2, g2 k Λ h Λ h, (2.5.68) w k+1 1 = g k γ k w 1 k, (2.5.69) w k+1 2 = g k γ k w k 2. (2.5.7) Then do m = m + 1 and go back to (2.5.54). 2.6 Numerical results Rotation of a single particle We have first considered the cases of a single neutrally buoyant ball in a bounded shear flow of a Newtonian fluid. The ball is placed at the middle between two walls initially, and it remains there in simulation even though it can move freely in fluid. The computational domain is = ( 1, 1) ( 1, 1) ( H/2, H/2) (i.e., L 1 = 2 and L 2 = 2) for different values of the height H. The ball radius is a =.15 and its mass center is located at (,,) initially. The blockage ratio is defined by K = 2a/H. 21

31 2.6. NUMERICAL RESULTS The shear rate γ = 1 so the velocity of the top wall is U = H/2 and that of the bottom wall is U = H/2. The fluid and particle densities are ρ f = ρ s = 1, the fluid viscosity being µ = 1. The mesh sizes for the velocity field is h = 1/48 or 1/64 and the mesh size for the pressure is 2h, the time step being t =.1. For all the numerical simulations considered in this section, we assume that all dimensional quantities are in the CGS units. 1.9 h = 1/48 h = 1/ ω K Under creeping conditions, the rotating velocity of the ball with respect to the x 2 -axis (see Fig.??) is γ/2 =.5 in an unbounded shear flow of a Newtonian fluid, according to the associated Jeffery s solution [1]. In Fig.??, the ball rotating 22

32 2.6. NUMERICAL RESULTS velocities have been shown for different values of the blockage ratio. Our numerical results are in a good agreement with the Jeffery s solution for most values of the blockage ratio; but we can observe the wall influence on the rotating velocity for the largest value of the blockage ratio, K =.3, in Fig.?? Sedimentation of a single particle In this section we have considered the terminal speed of sedimentary single particle in a vertical channel of infinitie length fillled with a Newtonian fluid. The computational domain is = ( 1, 1) ( 1, 1) ( 1, 1). The ball radius is a =.1 and its mass center is located at (,,) initially. The fluid and particle densities are ρ f = 1,ρ s = 1.5, the fluid viscosity being µ = 1. The mesh sizes for the velocity field is h = 1/48, 1/64, or 1/8 and the mesh size for the pressure is 2h, the time step being t =.1. We can valid the numerical results with the theoretical solution in [?, 7]. The formula of the theoretical solution of terminal speed V of sedimentary single particle is V = 2 ρ s ρ f ga 2 (2.6.71) 9 µ where g is the gravity. So the theoretical solution in our case is

33 2.6. NUMERICAL RESULTS Mesh Terminal Speed Size H Collocation Boundary Method Immersed Boundary Method 1/ / / Table

34 2.6. NUMERICAL RESULTS Two balls interaction in an one wall-driven bounded shear flow X 1 25

35 2.6. NUMERICAL RESULTS X Two balls interaction in a two wall-driven bounded shear flow X 1 26

36 27

37 3.1. SEVERAL MODELS OF NON-NEWTONIAN FLUID CHAPTER 3 3D DLM/FD methods for simulating the motion of spheres in bounded shear flows of Oldroyd-B fluids 3.1 Several Models of Non-Newtonian Fluid Stokes Equations Consider the Stokes equations in (, T ) describing the 3-dimensional motion of an incompressible and isothermal fluid without body force: σ = ρ f g, u =. 28

38 3.1. SEVERAL MODELS OF NON-NEWTONIAN FLUID where σ = pi + τ with presure p and stress tensor τ, u = (u 1, u 2, u 3 ) is the flow velocity, g is the gravity, and ρ f is the density of fluid Newtonian Model The constitutive equation of the stress tensor is given by τ = 2µD(u) where µ is the solvent viscosity of the fluid and D(u) is the symmetric rate of strain tensor 2D(u) = u + ( u) t, with the Jacobian of the velocity u The UCM-model In the UCM-model, we use the following constitutive equation to describe the viscoelastic stress tensor τ E : λ 1 τe + τ E = 2ηD(u). Here τ E is called the upper-convected time derivative of τ E and is defined by τe = τ E t + (u )τ E τ E ( u) t ( u) τ E, where λ 1 is the relaxation time of the fluid, and η is the elastic viscosity of the fluid. 29

39 3.1. SEVERAL MODELS OF NON-NEWTONIAN FLUID So we have the constitutive equation τ E t + (u )τ E τ E ( u) t ( u) τ E + 1 λ 1 τ E = 2η λ 1 D(u) The Oldroyd-B model In the more general Oldroyd-B model, the stress tensor τ is composed of a Newtonian component τ s and a viscoelastic component τ E, τ = τ s + τ E, and τ s is governed by a Newtonian consitutive equation τ s = 2µD(u). Thus, the Oldroyd-B model can be seen as a linear interpolation of the UCM and the Newtonian model: σ = ρ f g, u =, λ 1τE + τ E = 2ηD(u), where σ = pi + τ s + τ E. Using σ = pi + 2µD(u) + τ E, we obtain p 2µ D(u) τ E = ρ f g. Remark 3.1. Using the definition of conformation tenser C = λ 1 η τ E + I, we obtain 3

40 3.1. SEVERAL MODELS OF NON-NEWTONIAN FLUID the Oldroyd-B model in terms of C: p 2µ D(u) η (C I) = ρ f g, λ 1 u =, C t + (u )C ( u) C C ( u)t + 1 C = 1 I, λ 1 λ 1 where λ 1 is the relaxation time of the fluid, η is the elastic viscosity of the fluid, and I is the identity matrix. Here the conformation tensor C is symmetric and positive definite The Carreau model We also consider the non-newtonian fluid with shear thinning. In pure Oldroyd-B Model, the viscosity of fluid η 1 is a constant on the given domain and we define the elastic viscosity of the fluid η to be ( η = η 1 1 λ ) 2 λ 1 where λ 1 is relaxation time and λ 2 is retardation time. To consider the shear thinning in Oldroyd-B model, we use Carreau model to simulate it. Under Carreau model, the viscosity of fluid η 1 ( γ e ) depends on the fluid velocity. We define η 1 ( γ e ) to be η 1 ( γ e ) = η 1 ( ) 1 n 1 + (λ 1 γ e ) 2 2 (3.1.1) 31

41 3.2. DLM/FD METHOD FOR SIMULATING FLUID-PARTICLE INTERACTION IN STOKES FLOW where η 1 is the fluid viscosity without shear thinning, a number less than 1. γ e = 2D(u) : D(u) and n is 3.2 DLM/FD method for simulating fluid-particle interaction in stokes flow The governing equations Let be a bounded domain in R 3 and let Γ be the boundary of. We suppose that is filled with a viscoelasic fluid of Oldroyd-B type with density ρ f and contains N moving particles of density ρ s. Let B(t) = N i=1 B i (t) where B i (t) is the i-th solid particle in the fluid for i = 1, 2,, N. We denote by γ i (t) the boundary B i (t) of B i (t) for i = 1, 2,, N and let γ(t) = N i=1 γ i (t). For some T >, the governing equations for the fluid-particles system is as follows: For the motion of fluid, we consider the Stokes equations in Oldroyd-B model p 2µ D(u) η λ 1 (C I) = ρ f g in \ B(t), t (, T ), (3.2.2) u = g u = in \ B(t), t (, T ), (3.2.3) on Γ (, T ), with Γ g ndγ =, (3.2.4) u(x, t) = V i (t) + ω i (t) G i (t)x, x γ i (t), i = 1, 2,, N, (3.2.5) 32

42 3.2. DLM/FD METHOD FOR SIMULATING FLUID-PARTICLE INTERACTION IN STOKES FLOW C t + (u )C ( u) C C ( u)t + 1 λ 1 C = 1 λ 1 I in \ B(t), t (, T ), (3.2.6) C(x, ) = C (x), x \ B(), (3.2.7) C = C L on Γ, (3.2.8) where n is the unit normal vector pointing outward to the flow region. In (3.2.5), we assume a no-slip condition on the boundary of particles γ(t) u(x, t) = V i (t) + ω i (t) G i (t)x, x γ i (t), i = 1, 2,, N, where V i is the translation velocity, ω i is the angular velocity, G i is the center of mass and x is a point on the surface of the particle. The motion of particle satisfies the Euler-Newton s equations v i (x, t) = V i (t) + ω i (t) G i (t)x, {x, t} B i (t) (, T ), i = 1, 2,, N, (3.2.9) dg i dt = V i, (3.2.1) M i dv i dt d(i i ω i ) dt = M i g + F i + F r i, (3.2.11) = T i + G i x r F r i, (3.2.12) G i () = G i, V i () = V i, ω i () = ω i, (3.2.13) for i = 1, 2,, N, where M i and I i are the mass and the moment of inertia of the i-th particle, respectively; F i and T i are the hydrodynamical force and torque 33

43 3.2. DLM/FD METHOD FOR SIMULATING FLUID-PARTICLE INTERACTION IN STOKES FLOW imposed on the i-th particle by the fluid, and F r i is a short range repulsion force imposed on the i-th particle by other particles and wall to prevent particle/particle and particle/wall penetration. The hydrodynamical force F i and torque T i imposed on the i-th particle by the fluid are given by F i = σn dγ, γ T i = G i x σn dγ. (3.2.14) γ G i x r F r i is a torque acting on the point x r where F r i applies on the i-th particle DLM/FD formulation To obtain a distributed Lagrange multiplier/ fictitious domain formulation for the above problem (3.2.2) (3.2.14), we have the following three steps, namely: (i) we derive a global variational formulation of the virtual power type of problem (3.2.2) (3.2.14), (ii) we then extend the fluid motion into the region of particles with rigid body motion constraint, and then (iii) we relax that constraint by using a distributed lagrange multiplier. For convenience of derivation, we assume there is only one particle in the fluid, that is, we set B(t) the solid particle in the fluid, γ(t) the boundary B(t) of B(t), G(t) the center of mass of this particle. In the equations of the motion of particle, we set V the translation velocity of the particle B(t), ω the angular velocity of the particle B(t), M p and I p the mass and the moment of inertia of the particle B(t), 34

44 3.2. DLM/FD METHOD FOR SIMULATING FLUID-PARTICLE INTERACTION IN STOKES FLOW respectively; F and T the hydrodynamical force and torque imposed on the particle B(t) by the fluid, respectively, and F r a short range repulsion force imposed on the particle B(t). To obtain a variational formulation for above problem (3.2.2) (3.2.8), we define the following function spaces { W g (t) = v v (H 1 ( \ B(t))) 3, v = g (t) on Γ, v = V(t) + ω(t) G(t)x } on B(t), { W (t) = (v, Y, θ) v (H 1 ( \ B(t))) 3, v = on Γ, v = Y + θ G(t)x { L 2 ( \ B(t)) = q q L2 ( \ B(t)), \B(t) } q dx =, on B(t), with Y R and { } W( \ B(t)) = A A = [a ij ] M 3 3, a ij H 1 ( \ B(t)), i, j = 1, 2, 3, { } W CL ( \ B(t)) = A A W( \ B(t)), A = C L (t) on Γ. Applying the virtual power principle to the system (3.2.2) (3.2.8) obtains the following variational formulation: For a.e. t (, T ), find u(t) W g (t), p(t) L 2 ( \ B(t)), C(t) W CL,V(t) 35

45 3.2. DLM/FD METHOD FOR SIMULATING FLUID-PARTICLE INTERACTION IN STOKES FLOW R 3, G(t) R 3, ω(t) R 3, such that p vdx + 2µ f D(u) : D(v)dx \B(t) \B(t) η ( ) dv v ( (C I))dx + M p λ 1 \B(t) dt M pg F r Y ( d (Ip ω) + ) Gx r F r θ = ρ f g v dx (v, Y, θ) W (t), dt \B(t) (3.2.15) q u dx =, q L 2 ( \ B(t)), (3.2.16) \B(t) s W, \B(t) ( C t + (u )C ( u) C C ( u)t + 1 ) C : s dx = 1 I : s dx, λ 1 λ 1 \B(t) dg dt (3.2.17) = V, (3.2.18) u(x, ) = u (x), x \ B(), (3.2.19) C(x, ) = C (x), x, (3.2.2) G(x, ) = G (x), V(x, ) = V (x), ω(x, ) = ω (x), (3.2.21) To obtain an equivalent fictitious domain formulation, first we fill B with a fluid of density ρ f and suppose that this fluid follows the same rigid body motion as B itself, which is u(x, t) = V(t) + ω(t) G(t)x, x B(t). (3.2.22) 36

46 3.2. DLM/FD METHOD FOR SIMULATING FLUID-PARTICLE INTERACTION IN STOKES FLOW Define a function space { W (t) = (v, Y, θ) (v \B(t), Y, θ) W (t), v(x, t) = Y + θ G(t)x } x B(t). Suppose particle B is made of an homogeneous material of density ρ f which follows ρ f B(t) g v dx = ρ f ρ s M p g Y, (v, Y, θ) W (t), (3.2.23) v = in B(t), (v, Y, θ) W (t), (3.2.24) u = in B(t) and D(u) = in B(t). (3.2.25) To obtain a fictitious domain formulation, we define the following function spaces and V g (t) = { v v (H 1 ()) 3, v = g (t) on Γ }, { L 2 () = q q L2 (), } q dx =, V CL () = { A A W(), A = CL (t) on Γ }, V C () = { A A W(), A = on Γ }. Combining (3.2.15)-(3.2.21) with (3.2.22)-(3.2.25), we obtain the fictitious domain formulation as follows: For a.e. t (, T ), find u(t) V g (t), p(t) L 2 (), C(t) V CL,V(t) R 3, 37

47 3.2. DLM/FD METHOD FOR SIMULATING FLUID-PARTICLE INTERACTION IN STOKES FLOW G(t) R 3, ω(t) R 3, such that (p =, D(u) =, C = I, u = in B(t) ) p v dx + 2µ f η λ 1 D(u) : D(v) dx dv v ( (C I)) dx + M p dt Y + d (I pω) dt ( Gx r F r θ = ρ f g v dx + 1 ρ f ρ s F r Y (v, Y, θ) W (t), θ ) M p g Y, (3.2.26) q u dx =, q L 2 (), (3.2.27) ( C t + (u )C ( u) C C ( u)t + 1 ) C λ 1 s V C, C = I in B(t), dg dt : s dx = 1 λ 1 I : s dx, (3.2.28) = V, (3.2.29) u(x, t) = V(t) + ω(t) G(t)x, x B(t). (3.2.3) u(x, ) = u (x), x \ B(), (3.2.31) C(x, ) = C (x), x, (3.2.32) G(x, ) = G (x), V(x, ) = V (x), ω(x, ) = ω (x), (3.2.33) To relax the rigid body motion condition (3.2.3), we introduce a Lagrange multiplier, λ Λ(t) = (H 1 (B(t))) 3, and for any µ (H 1 (B(t))) 3 and v (H 1 ()) 3 38

48 3.2. DLM/FD METHOD FOR SIMULATING FLUID-PARTICLE INTERACTION IN STOKES FLOW such that µ, v Λ(t) = (µ v + 1d ) µ : v dx, 2 B(t) where d is a scaling constant and, typically, has been used as the diameter of the particles. Then we obtain the fictitious domain formulation over the entire region as follows: For a.e. t (, T ), find u(t) V g (t), p(t) L 2 (), C(t) V CL,V(t) R 3, G(t) R 3, ω(t) R 3, λ(t) Λ(t) such that dv +M p dt Y + d (I pω) θ F r Y Gx r F r θ dt λ, v Y θ Gx = ρ f g v dx + p v dx + 2µ f D(u) : D(v) dx η λ 1 Λ(t) (v, Y, θ) (H 1 ()) 3 R 3 R 3, v ( (C I)) dx ( 1 ρ f ρ s ) M p g Y, (3.2.34) q u dx =, q L 2 (), (3.2.35) ( C t + (u )C ( u) C C ( u)t + 1 ) C λ 1 s V C, C = I in B(t), dg dt : s dx = 1 λ 1 I : s dx, (3.2.36) = V, (3.2.37) u(x, ) = u (x), x \ B(), (3.2.38) 39

49 3.3. NUMERICAL METHODS µ, u(t) V(t) ω(t) G(t)x =, µ Λ(t), (3.2.39) Λ(t) C(x, ) = C (x), x, (3.2.4) G(x, ) = G (x), V(x, ) = V (x), ω(x, ) = ω (x). (3.2.41) Remark 3.2. Since u is divergence free and satisfies the Dirichlet boundary conditions on Γ, we obtain 2 D(u) : D(v) dx = u : v dx, v (H()) 1 3. So in relation (3.2.34) we can replace 2 D(u) : D(v) dx by u : v dx. Also the gravity g in (3.2.34) can be absorbed into the presure term. 3.3 Numerical methods Finite element approximation For the purpose of finding an approximation solution of problem (3.2.34)-(3.2.41), we need a partition of the flow region R 3. We use an uniform finite element mesh for and let h be the uniform finite element mesh size for the velocity field, T h be a tetrahedrization of, and T 2h be twice coarser than T h. For the space discretization, let P i be the space of the polynomials in three variables of degree i, we have chosen P 1 -iso-p 2 finite element space for the velocity field and conformation tensor and P 1 finite elementspace for the pressure. Then we 4

50 3.3. NUMERICAL METHODS have the following function spaces: V h = {v h vh ( C ( ) ) 3, vh T (P 1 ) 3, T T h }, V gh (t) = {v h v h V h, v h Γ = g h (t) }, V h = {v h v h V h, v h Γ = }, and V CLh = V Ch = L 2 h = { q h qh C ( ), q h T P 1, T T 2h }, L 2 h = { } q h q h L 2 h, q h dx =, { sh s h ( C ( ) ) 3 3 T, sh (P 1 ) 3 3 Γ, T T h, s h = C Lh (t) h { sh s h ( C ( ) ) } 3 3 T, sh (P 1 ) 3 3 Γ, T T h, s h =, h }, where g h (t) is an approximation of g (t) satisfying g h (t) ndγ =, Γ and Γ h = {x x Γ, g h(x, t) n(x) < }. For simulating the particle motion in fluid flow, we need to define a finite dimensional space to approach the space of Lagrange multipier Λ(t). Let {ξ i } N(t) i=1 be a set of points from B(t) which cover B(t) evenly. We define the discrete Lagrange multiplier space 41

51 3.3. NUMERICAL METHODS by N(t) Λ h (t) = µ h µ h = µ i δ(x ξ i ), µ i R 3, i = 1,, N(t), i=1 where x δ(x ξ i ) is the Dirac measure at x = ξ i. Then we define a pairing over Λ h (t) V gh (t) (or Λ h (t) V g ) by N µ h, v h Λh (t) = µ i v h (ξ i ), (3.3.42) i=1 for µ h Λ h (t), v h V gh (t) or V h. A typical set {ξ i } N(t) i=1 of points from B(t) to be used in (3.3.42) is defined as {ξ i } N(t) i=1 = {ξ i} N 1(t) i=1 {ξi } N(t) i=n 1 (t)+1, where {ξ i } N 1(t) i=1 is the set of those vertices of the velocity grid T h contained in B(t) and the distance between those vertices and the boundary B(t) is greater than or equal to h 2, and selected points {ξ i} N(t) i=n 1 (t)+1 from B(t). For simulating particle interactions in Stokes flow, we define a modified pairing <, > Λh (t) as follows: N 1 (t) µ h, v h Λh (t) = µ i v h (ξ i ) + i=1 N(t) i=n 1 (t)+1 j=1 M µ i v h (ξ i )D h (ξ i x j )h 3, (3.3.43) for µ h Λ h (t), v h V gh (t) or V h where {x j } M j=1 are the grid points of the finite elements for the velocity, and the function D h (X ξ i ) is defined as D h (X ξ i ) = δ h (X 1 ξ i1 )δ h (X 2 ξ i2 )δ h (X 3 ξ i3 ), 42

52 3.3. NUMERICAL METHODS with X = (X 1, X 2, X 3 ) t and ξ i = (ξ i1, ξ i2, ξ i3 ) t, and the one-dimensional approximate Dirac measure δ h being defined by δ h (z) = z 8h h z h z 8h h z h 4 ( ) 2 z, z h, h ( ) 2 z, h z 2h, h, otherwise. By using the above pairing <, > Λh (t), the rigid body motion of particle B(t) is forced via an immersed boundary method develped by Peskin. Based on the finite dimensional spaces above, we obtain the following approximation of problem (3.2.34)- (3.2.41) : For a.e. t (, T ), find u h (t) V gh (t), p(t) L 2 h, C h(t) V CLh (t),v(t) R 3, G(t) R 3, ω(t) R 3, λ h Λ h (t) such that p v dx + µ f F r Y Gx r F r θ = (v, Y, θ) V h R 3 R 3, u h : v dx η dv v ( (C h I)) dx + M p λ 1 dt Y + d (I pω) θ dt ( 1 ρ ) f M p g Y + λ h, v Y θ Gx, ρ s Λ h (t) (3.3.44) q u h dx =, q L 2 h, (3.3.45) 43

53 3.3. NUMERICAL METHODS ( Ch t + (u h )C h ( u h ) C h C h ( u h ) t + 1 ) C h : s h dx = 1 λ 1 λ 1 s h V Ch, C h = I in B h (t), dg dt (3.3.46) = V, (3.3.47) I : s h dx, u h (x, ) = u h (x), x \ B h (), (3.3.48) µ h, u h (t) V(t) ω(t) G(t)x =, µ h Λ h (t), (3.3.49) Λ h (t) C h (x, ) = C h (x), x, (3.3.5) G(x, ) = G (x), V(x, ) = V (x), ω(x, ) = ω (x), (3.3.51) where u h is an approximation of u such that q u h dx =, q L 2 h. Remark 3.3. If we consider our particle as a sphere, the moment of inertia I p is a constant with respect to time t. Then in relation (3.3.44) the term d (I pω) can be dt dω rewriten as I p dt Operator splitting scheme Consider the following initial value problem: dφ + A(φ) = on (, T ), dt φ() = φ, 44

54 3.3. NUMERICAL METHODS with < T < +. A = J A j with J 2. j=1 We suppose that operator A has a decompsition such as Let τ > be a time-discretization step, we denote nτ by t n. By setting φ n as an approximation of φ(t n ), we can write down the Lie s scheme as follows: Given φ = φ. For n, φ n is known and we compute φ n+1 via dφ dt + A j(φ) = on (t n, t n+1 ), φ(t n j 1 n+ ) = φ J ; φ n+ j J = φ(t n+1 ), for j = 1,, J. Since the conformation tensor C is symmetric and positive definite, by Cholesky factorization there is a 3 3 lower triangular matrix A such that C = AA t, where A t is the transpose matrix of A. To split the constitutive equation with respect to C, we have derived the following results. Lemma 3.4. For a lower triangular matrix A and C = AA t, given u R 3 and a positvie constant λ 1, we have (i) if A satisfies the equation da + (u )A =, then C satisfies the equation dt dc + (u )C = ; dt (ii) if A satisfies the equation da dt + 1 2λ 1 A ( u)a =, then C satisfies the equation dc dt + 1 λ 1 C ( u)c C( u) t =. Proof. (i) Given da dt + (u )A =. Mulitplying the equation by At to the right, 45

55 3.3. NUMERICAL METHODS we obtain da dt At + (u )AA t =. (3.3.52) Mulitplying the transpose of the equation by A to the left, we obtain A dat dt + A(u )A t =. (3.3.53) Adding (3.3.52) and (3.3.53) gives d (AA t ) dt + (u ) ( AA t) =. Thus, we get dc dt + (u )C =. (ii) Given da dt + 1 A ( u)a =. Mulitplying the equation by A t to the right, 2λ 1 we obtain da dt At + 1 AA t ( u)aa t = (3.3.54) 2λ 1 Mulitplying the transpose of the equation by A to the left, we obtain A dat dt + 1 2λ 1 AA t AA t ( u) t = (3.3.55) Adding (3.3.54) and (3.3.55) gives d (AA t ) dt + 1 λ 1 AA t ( u)aa t AA t ( u) t =. 46

56 3.3. NUMERICAL METHODS Thus, we get dc dt + 1 λ 1 C ( u)c C( u) t =. Similarly, we can define finite dimensional spaces V ALh (t) or V Ah for A. Applying the Lie s scheme to the discrete analogue of the problem (3.3.44)- (3.3.51) with C = AA t and the backward Euler s method to some subproblems, we obtain: Given u = u h, C = C h, G = G, V = V, ω = ω. For n, u n, C n, G n, V n, ω n are known, we first predict the position and the translation velocity of the center of mass as follows. dg = V(t), dt (3.3.56) dv M p =, dt (3.3.57) dω I p =, dt (3.3.58) V(t n ) = V n, ω(t n ) = ω n, G(t n ) = G n, (3.3.59) for t n < t < t n+1. Then set V n+ 1 4 = V(t n+1 ), ω n+ 1 4 = ω(t n+1 ), and G n+ 1 4 = G(t n+1 ). With the center G n+1 we get in the above step, the region of B n+1 h occupied by the particle is determined and we set C n+ 1 4 = I in B n+1 h and C n+ 1 4 = C n otherwise. Then we enforce the rigid body motion in B n+1 h and solve u n+1 and p n+1 simultaneously as follows: 47

57 3.3. NUMERICAL METHODS Find u n+1 Vg n+1 h, p n+1 L 2 h, λn+1 Λ n+1, V n+1 R 3, ω n+1 R 3 such that p n+1 v dx + µ f u n+1 : v dx η λ 1 V n+1 V n+ 1 4 ω n+1 ω n M p Y + I p θ t t ( = 1 ρ ) f M p g Y + λ n+1, v Y θ G n+1 x ρ s (v, Y, θ) V h R 3 R 3, µ, u n+1 V n+1 ω n+1 G n+1 x and set C n+ 2 4 = C n h v ( )) (C n+ 1 4 I dx Λ n+1 h, (3.3.6) q u n+1 dx =, q L 2 h, (3.3.61) Λ n+1 h =, µ Λ n+1 h. (3.3.62) Next, we set A n+ 2 4 (A n+ 2 4 ) t = C n+ 2 4 and compute A n+ 3 4 via the solution of da(t) dt : s dx + (u n ) A(t) : s dx =, s V Ah A(t n ) = A n+ 2 4, A(t) V n+1 A Lh, t [t n, t n+1 ] (3.3.63) and set A(t n+1 ) = A n+ 3 4 and u n+ 3 4 = u n Then we compute A n+1 via the solution of [ A n+1 A n+ 3 4 t ] ( u n ) A n A n+1 : s dx =, 2λ 1 s V Ah, A n+1 V n+1 A Lh, (3.3.64) 48

58 3.3. NUMERICAL METHODS and set C n+1 = A n+1 (A n+1 ) t + 1 λ 1 I. (3.3.65) In the above,v n+1 A Lh = V ALh (t n+1 ), V n+1 g h = V gh (t n+1 ), Λ n+1 h = Λ h (t n+1 ), and B n+s h = B h (t n+s ) where the spaces V ALh (t) and V Ah for A are defined similar to those V CLh (t) and V Ch Logarithm of conformation tensor Besides matrix factorization approach, we consider the log-conformation representation for the conformation tensor. add the derivation of logarithm conformation. In order to resolve the exponential behavior of the conformation tensor, we replace (3.3.63) (3.3.65) by (3.3.66) (3.3.68) when solving the constitutive equation: We set ψ n+ 2 4 ( ) = log C n+ 2 4 and compute ψ n+ 3 4 via the solution of ψ(t) t : s dx + (u n ) ψ(t) : s dx =, s V Ch ψ(t n ) = ψ n+ 2 4, ψ(t) V n+1 C Lh, t [t n, t n+1 ] (3.3.66) and set ψ n+ 3 4 = ψ(t n+1 ) and u n+ 3 4 = u n

59 3.3. NUMERICAL METHODS Then, we set C n+ 3 4 = e ψn+ 3 4 and O n S n N n+ 3 4 compute ψ n+1 via the solution of ( ) 1 C n+ 3 4 = u n+ 3 4 and ψ(t) t ( ) O n+ 3 4 ψ(t) ψ(t)o n S n+ 3 4 =, ψ(t n ) = ψ n+ 3 4, ψ(t) V n+1 C Lh, t [t n, t n+1 ] (3.3.67) and set ψ n+1 = ψ(t n+1 ) and C n+1 = e ψn+1. Then solve the ODE C t = 1 (I C), λ 1 C(t n ) = C n+1, C(t) V n+1 C Lh, t [t n, t n+1 ] (3.3.68) In (3.3.67), it is better to use exact solution instead of the approxiamtion from the first order scheme Operator splitting scheme of Carreau model To consider the operator splitting scheme of Carreau model, we have: Given u = u h, C = C h, G = G, V = V, ω = ω. For n, u n, C n, G n, V n, ω n are known, we first predict the position and the translation velocity of the center of mass as follows. Taking V n+ 1 4, = V n and G n+1, = G n, for k = 1, 2,, N, 5

60 3.3. NUMERICAL METHODS computing dg = V(t), dt (3.3.69) dv M p =, dt (3.3.7) dω I p =, dt (3.3.71) V(t n ) = V n, ω(t n ) = ω n, G(t n ) = G n, (3.3.72) for t n < t < t n+1.then set V n+ 1 4 = V(t n+1 ), ω n+ 1 4 = ω(t n+1 ), and G n+ 1 4 = G(t n+1 ). With the center G n+1 we get in the above step, the region of B n+1 h occupied by the particle is determined and we set C n+ 1 4 = I in B n+1 h and C n+ 1 4 = C n otherwise. Then we calculate η 1 ( γ n ) by using the previous time step velocity u n and set µ( γ n ) = η 1 ( γ n ) λ 2 B n+1 h λ 1 and η( γ n ) = η 1 ( γ n ) µ( γ n ), enforce the rigid body motion in and solve u n+1 and p n+1 simultaneously as follows: Find u n+1 Vg n+1 h, p n+1 L 2 h, λn+1 Λ n+1, V n+1 R 3, ω n+1 R 3 such that h ( γn ) 2 = 2D(u n ) : D(u n ), η 1 ( γn ) = η 1 ( 1 + (λ 1 γn ) 2 ) 1 n 2 ( ) ( ) λ 2, µ γn = η 1 γn, λ 1 ( ) ( ) ( ) η γn = η 1 γn µ γn, 51

61 3.3. NUMERICAL METHODS p n+1 v dx + µ f u n+1 V n+1 V n+ 1 4 ω n+1 ω n+ 1 4 : v dx + M p Y + I p t t ( ( )) ( ) η γn ( ) = 2 µ f µ γn (D(u n ) : v) dx + v C n+ 1 4 I dx λ 1 ( + 1 ρ ) f M p g Y + λ n+1, v Y θ G n+1 x, ρ s (v, Y, θ) V h R 3 R 3, µ, u n+1 V n+1 ω n+1 G n+1 x Λ n+1 h (3.3.73) q u n+1 dx =, q L 2 h, (3.3.74) Λ n+1 h =, µ Λ n+1 h. (3.3.75) θ and set C n+ 2 4 = C n Next, we set A n+ 2 4 (A n+ 2 4 ) t = C n+ 2 4 and compute A n+ 3 4 via the solution of da(t) dt : s dx + (u n ) A(t) : s dx =, s V Ah A(t n ) = A n+ 2 4, A(t) V n+1 A Lh, t [t n, t n+1 ] (3.3.76) and set A(t n+1 ) = A n+ 3 4 and u n+ 3 4 = u n Then we compute A n+1 via the solution of [ A n+1 A n+ 3 4 t ] ( u n ) A n A n+1 : s dx =, 2λ 1 s V Ah, A n+1 V n+1 A Lh, (3.3.77) 52

62 3.4. ON THE SOLUTION OF THE SUBPROBLEMS FROM OPERATOR SPLITTING and set C n+1 = A n+1 (A n+1 ) t + 1 λ 1 I. (3.3.78) In the above,v n+1 A Lh = V ALh (t n+1 ), V n+1 g h = V gh (t n+1 ), Λ n+1 h = Λ h (t n+1 ), and B n+s h = B h (t n+s ) where the spaces V ALh (t) and V Ah for A are defined similar to those V CLh (t) and V Ch. 3.4 On the solution of the subproblems from operator splitting Solution of the advection subproblems We solve the advection problem (3.3.63) by a wave-like equation method [4]. After translation and dilation on the time axis, each component of the velocity vector u and of the tensor A is solution of a transport equation of the following type: ϕ t + (U )ϕ = in (tn, t n+1 ), ϕ() = ϕ, ϕ = g on Γ (t n, t n+1 ), (3.4.79) 53

63 3.4. ON THE SOLUTION OF THE SUBPROBLEMS FROM OPERATOR SPLITTING where U = and U t well-posed problem: = on (, t). Thus, (3.4.79) is equivalent to the 2 ϕ t 2 ϕ() = ϕ, (U n) ((U ϕ)u) = in (, t), ( ϕ t ϕ t () = U ϕ, ϕ = g on Γ (t n, t n+1 ), ) + (U )ϕ = on Γ \ Γ (t n, t n+1 ). (3.4.8) Solving the wave-like equation (3.4.8) by a classical finite element/ time stepping method, a variational formulation of (3.4.8) is given by 2 ϕ t vdx + (U ϕ)(u v)dx + U n ϕ 2 Γ\Γ t vdγ =, v W, ϕ() = ϕ, ϕ t () = U ϕ, ϕ = g on Γ (t n, t n+1 ), with the test function space W defined by W = {v v H 1 (), v = on Γ }. (3.4.81) Let H 1 h be a C -conforming finite element subspace of H 1 (). We define W h = Hh 1 W. We suppose that lim W h = W in the usual finite element sense. Next, we h define τ 1 > by τ 1 = t where Q is a positive integer, and we discretize problem Q (3.4.81) by ϕ = ϕ, (ϕ 1 ϕ 1 )vdx = 2τ 1 (U h ϕ )vdx, v W h, ϕ 1 ϕ 1 W h, (3.4.82) 54

64 3.4. ON THE SOLUTION OF THE SUBPROBLEMS FROM OPERATOR SPLITTING and for q =, 1, 2,, Q 1, ϕ q+1 Hh 1, ϕq+1 = g h on Γ, ϕ q+1 + ϕ q 1 2ϕ q vdx + (U τ 2 h ϕ q )(U h v)dx ( ) ϕ q+1 ϕ q 1 + U h n vdγ =, v W, Γ\Γ τ (3.4.83) where U h and g h are the approximates of U and g, respectively. Remark 3.5. Scheme (3.4.82) (3.4.83) is a centreed scheme which is formally second order accurate with respect to space and time discretizations. To be stable, scheme (3.4.82) (3.4.83) has to verify a condition such as τ 1 ch, which c of 1 order of U. Since the advection problem is decoupled from the other ones, we can choose proper time step here so that the above condition is satisfied. If one uses the trapezoidal rule to compute the first and the third integrals in (3.4.83), the above scheme becomes explicit and ϕ q+1 is obtained via the solution of a linear system with diagonal matrix. Remark 3.6. Scheme (3.4.82) (3.4.83) does not introduce numerical dissipation, unlike the upwinding schemes commonly used to solve transport problems like (3.4.79). Remark 3.7. If we consider the homogeneous boundary condition, U h Γ = and set Q = 1 in (3.4.82) (3.4.83). Then we obtain the following: ϕ 1 ϕ vdx + (U h ϕ )vdx t = t (U h ϕ )(U h v)dx, v H 1 2 h; ϕ Hh 1 (3.4.84) 55

65 3.4. ON THE SOLUTION OF THE SUBPROBLEMS FROM OPERATOR SPLITTING Solution of the rigid body motion enforcement problems In system (3.3.6)-(3.3.62), there are two multipliers. p and λ. We have solved this system via an Uzawa-conjugate gradient method driven by both multipliers. The general problem is as follows: Find u V gh, p L 2 h, λ Λ h, V R 3, ω R 3 such that p v dx + µ f u : v dx η λ 1 V V ω ω +M p Y + I p θ t t ( = 1 ρ ) f M p g Y + λ, v Y θ Gx ρ s (v, Y, θ) V h R 3 R 3, v ( (C I)) dx Λ h, (3.4.85) q u dx =, q L 2 h, (3.4.86) µ, u V ω Gx =, µ Λ h. (3.4.87) Λ h Applying the following Uzawa-conjugate gradient algorithm operating in the space L 2 h Λ h to solve the sysyem (3.4.85)-(3.4.87): Assume p L 2 h and λ Λ h are given. We solve the problem: Find u V gh, V R 3, ω R 3 satisfying 56

66 3.4. ON THE SOLUTION OF THE SUBPROBLEMS FROM OPERATOR SPLITTING µ f u : v dx = v V h ; u V gh, M p V V t and then compute p v dx + η λ 1 v ( (C I)) dx + λ, v Λ h, (3.4.88) ( Y = 1 ρ ) f M p g Y λ, Y, Y R 3, (3.4.89) ρ Λ s h I p ω ω t θ = λ, θ Gx Λ h, θ R 3, (3.4.9) g 1 = u ; (3.4.91) next find g 2 Λ h satisfying µ, g 2 Λ h = µ, u V ω Gx, µ Λ h, (3.4.92) Λ h and set w 1 = g 1, w 2 = g 2. (3.4.93) Then for k, assuming that p k, λ k, u k, V k, ω k, g k 1, g k 2, w k 1 and w k 2 are known, compute p k+1, λ k+1, u k+1, V k+1, ω k+1, g k+1 1, g k+1 2, w 1 k+1 and w k+1 2 as follows: µ f u k : v dx = w1 k v dx + w2, k v, Λ h v V h ; u k V gh, (3.4.94) M p V k t Y = w k 2, Y Λ h, Y R 3, (3.4.95) 57

67 3.4. ON THE SOLUTION OF THE SUBPROBLEMS FROM OPERATOR SPLITTING and then compute ω k I p t θ = w2, k θ Gx, θ R 3, (3.4.96) Λ h g k 1 = u k ; (3.4.97) next find g k 2 Λ h satisfying µ, g k 2 Λ h = µ, u k V k ω k Gx, µ Λ h, (3.4.98) Λ h and compute and ρ k = g k 1 2 dx + g k 2, g2 k Λ h g k 1w1dx k +, (3.4.99) g k 2, w2 k Λ h p k+1 = p k ρ k w 1 k, (3.4.1) λ k+1 = λ k ρ k w k 2, (3.4.11) u k+1 = u k ρ k u k, (3.4.12) V k+1 = V k ρ k V k, (3.4.13) ω k+1 = ω k ρ k ω k, (3.4.14) g k+1 1 = g k 1 ρ k g k 1, (3.4.15) g k+1 2 = g k 2 ρ k g k 2. (3.4.16) 58

68 3.5. NUMERICAL RESULTS If g k dx + g2 k+1, g2 k+1 g 1 2 dx + g 2, g2 Λ h Λ h ε, (3.4.17) then take p = p k+1, λ = λ k+1, u = u k+1, V = V k+1, and ω = ω k+1. Otherwise, compute and set γ k = g k dx + g2 k+1, g2 k+1 g k 1 2 dx + g k 2, g2 k Λ h Λ h, (3.4.18) w k+1 1 = g k γ k w 1 k, (3.4.19) w k+1 2 = g k γ k w k 2. (3.4.11) Then do m = m + 1 and go back to (3.4.94). 3.5 Numerical results Rotation of a single particle We have considered the cases of a single neutrally buoyant ball placed at the middle between two walls initially with respect to relaxation time λ 1 in a bounded shear flow of Oldroyd-B fluids. The densities of the fluid and that of the particle are ρ f = ρ s = 1 and the viscosity µ f = 1. The computational domain is = ( x 1, x 1 ) ( x 2, x 2 ) ( x 3, x 3 )(i.e.,l 1 = 2x 1, L 2 = 2x 2, and L 3 = 2x 3 ).The shear rate γ = 1 sec 1 so the velocity of the top wall is U = x 3 and the velocity of the bottom wall is U = x 3. 59

69 3.5. NUMERICAL RESULTS For all the numerical simulations, we assume that all dimensional quantities are in the CGS units. We have obtained the rotating angular velocity with respect to the x 2 -axis for different values of λ 1 with the retardation time λ 2 = λ 1 /8. First, to study the slowing down effect in the particle rotating angular velocity due to a finite wall-particle distance, we consider the mass center of the ball is fixed at (,, ) with three different λ 1 and define the blockage ratio K = 2r/L 3 with five particle radii r = 1/1, 1/5, 1/3, 2/5, 1/2. The associated values of the Weissenberg number Wi ( = λ 1 γ) are.5,.75, and 1.. In order to assure the unpertubed conditions, the computational domain is = ( L/2, L/2) ( L/2, L/2) ( 1, 1) where L = 2r and L 3 = 2. So we can consider five different block ratios K as same as those cases in [5] for the numerical results. graph???? Second, to study the effect of viscoelasticity in the particle rotating angular velocity, we consider two different setups of mass center. First one is the cases of particle with fixed mass center at (,, ) all the time and the second one is the case of free moving particle without fixed mass center. The particle radius r is.1. The computational domain is = ( 1.5, 1.5) ( 1.5, 1.5) ( 1.5, 1.5). The mass center of the ball is located at (,, ) initially. We have obtained the rotating angular velocity with respect to the x 2 -axis for ten different values of the relaxation time λ 1 with the retardation time λ 2 = λ 1 /8. The associated values of the Weissenberg number Wi (= λ 1 γ) are.1,.1,.25,.5, 1, 1.6, 2.6, 3.56, 4.2, and

70 3.5. NUMERICAL RESULTS.5.4 Angular Velocity.3.2 Wi=.1 Wi=.1 Wi=.25 Wi=.5 Wi=1. Wi=1.6 Wi=2.6 Wi=3.56 Wi=4.2 Wi= Time 61

71 3.5. NUMERICAL RESULTS Migration of a single particle in an one wall driven bounded shear flow Two ball interacting with large initial distance in a two wall driven bounded shear flow.5 Z-axis X-axis Y-axis In this section we consider the cases of two balls of the same size interacting in a bounded shear flow fluid of Oldroyd-B type as visualized in Fig.??. The ball radii are r =.1. The fluid and ball densities are ρ f = ρ s = 1, the viscosity being µ = 1. The computational domain is = ( 1.5, 1.5) ( 1, 1) (.5,.5) (i.e., L 1 = 3, L 2 = 2, and L 3 = 1). The shear rate is fixed at γ = 1 so the velocity of the top wall is U =.5, the bottom wall being U =.5. The mass centers of the two balls are located on the shear plane at ( d,, s) and (d,, s) initially, where s varies and d is.5. The time step being t =.1. Then we consider six or seven 62

72 3.5. NUMERICAL RESULTS dimensionless initial vertical displacements D = s/a from the ball center to the middle plane in the two wall driven bounded shear flow..2.1 X X 1 When two balls move in a bounded shear flow of a Newtonian fluid at Stokes regime with D =.122,.194,.255,.316,.5, 1 as in Fig.??, the higher ball takes over the lower one and then both return to their initial heights for those large vertical displacements D =.316,.5 and 1. These two particle paths are called pass (or open) trajectories. But for smaller vertical displacements, D =.122,.255 and.316, they first come close to each other and to the mid-plane between the two horizontal walls, then, the balls move away from each other and from the above mid-plane. These two particle paths are called return trajectories. Both kinds are on the shear plane as shown in Fig.?? for Wi= (Newtonian case) and they are consistent with the results obtained in [21]. 63

73 3.5. NUMERICAL RESULTS.2 X Wi=.1 D = 1. D =.5 D =.42 D =.316 D =.255 D =.194 D = X 1.2 X Wi=.25 D = 1. D =.5 D =.42 D =.316 D =.255 D =.194 D = X 1.2 X Wi=.5 D = 1. D =.5 D =.42 D =.316 D =.255 D =.194 D = X 1 64

74 3.5. NUMERICAL RESULTS.2.1 X X 1.2 X Wi=1. D = 1. D =.5 D =.38 D =.316 D =.255 D =.194 D = X 1 For the two balls interacting in an Oldroyd-B fluid with D =.122,.194,.255,.316,.42,.5, 1, we have summarized the results for Wi=.1,.25,.5,.75, and 1 in Figs.?? to??. As in Newtonian fluids, there are results of pass and return trajectories concerning two ball encounters; but the those of the two ball mass centers lose the symmetry due to the effect of elastic force arising from viscoelastic fluids. For example, the open trajectories associated with D =.5 for Wi=.1,.25,.5, 65

75 3.5. NUMERICAL RESULTS and 1 are closer to the mid-plane after two balls pass over/under each other. The elastic force is not strong enough to hold them together during passing over/under, but it already pulls the balls toward each other and then change the shape of the trajectories. For higher values of Wi considered in this section, there are less return trajectories; instead it is easier to obtain the two ball chain once they run into each other. Actually depending on the Weissenberg number Wi and the initial vertical displacement s, a chain of two balls can be formed in a bounded shear flow, and then such chain tumbles. For example, for D =.316, the two balls come to each other, form a chain and then rotate with respect to the midpoint between two mass centers for Wi=.1,.25,.5, and 1. The details of the phase diagram of pass, return, and tumbling are shown in Fig.??. The range of the vertical distance for the passing over becomes bigger for higher Weissenberg numbers. For the shear flow considered in this article, the increasing of the value of the Wi with a fixed shear rate is same as to increase the shear rate with a fixed relaxation time. This explains why, for Wi=1, two balls can have bigger gap between them while rotating with respect to the middle point between two mass centers since the two balls are kind of moving under higher shear rate. Those tumbling trajectories are associated with the closed streamlines around a freely rotating ball centred at the origin. 66

76 3.5. NUMERICAL RESULTS 1.2 Passing Tumbling Swapping 1.8 D Wi 67

77 3.5. NUMERICAL RESULTS Two ball interacting with small initial distance in an one wall driven bounded shear flow.5 Z-Axis X-Axis Y-Axis In this section we consider the cases of two balls of the same size interacting in a bounded shear flow driven by the upper wall as visualized in Fig.??. The ball radii are r =.1. The fluid and ball densities are ρ f = ρ s = 1, the viscosity being µ = 1. The computational domain is = ( 1.5, 1.5) ( 1, 1) (.5,.5). The shear rate is fixed at γ = 1 but the velocity of the top wall is U = 1, the bottom wall being U =. The mesh size for the velocity field and the conformation tensor is h = 1/48, the mesh size for the pressure is 2h, The time step being t =.1. The mass centers of the two balls are located on the shear plane at ( x,, z ) and (x,, z ) initially such that the angle between the mid-plane and the line segment of two initial locations of mass center of particle is 175 counterclockwisely. We define the gap size 68

78 3.5. NUMERICAL RESULTS = d 2r where d is the distance between two mass centers of particle and r is X X 1 For the two balls interacting in Oldroyd-B fluid with gap= h/8, h/4, h/2, h, 2h, and 3h where h is the mesh size, we have summarized the results for Wi=.1,.25,.5,.75, and 1 in Figs.?? to??..2 X Wi=.25 gap= 3H gap= 2H gap= H gap= H/2 gap= H/4 gap= H/ X 1 69

79 3.5. NUMERICAL RESULTS.2 X Wi=.5 gap= 3H gap= 2H gap= H gap= H/2 gap= H/4 gap= H/ X X X X X 1 7

80 3.5. NUMERICAL RESULTS X X 1 71

81 3.5. NUMERICAL RESULTS Tumbling and Kayaking Tumbling Non-Tumbling 2.5 Gap Wi 72

82 3.5. NUMERICAL RESULTS.5.5 t=32 t=34 X 3 X X X t=36 t=38 X 3 X X X t=4 t=42 X 3 X X 1 t= X 1 t=46 X 3 X X X t=48 t=5 X 3 X X 1 t= X 1.5 t=54 X 3 X X X 1 Figure 3.5.1: The tumbling motion as W i =.5: the initial distance between two particles is 2r+gap where r =.1, gap=3h.for each frame, the horizontal direction is X 1 axis and vertical direction is X 3 axis. 73

83 3.5. NUMERICAL RESULTS.5.5 t=5 t=52 X 3 X X 2 X t=54 t=58.5 X 3 X X 2.5 X 2 t=59.5 t=61 X 3 X X 2.5 X 2 t=64 t=67 X 3 X X 2.5 X 2 t=68 t=69 X 3 X X 2 Figure 3.5.2: The kayaking motion as W i = 1.: the initial distance between two particles is 2r+gap where r =.1, gap=3h. For each frame, the horizontal direction is X 2 axis and vertical direction is X 3 axis. 74 X 2

84 CHAPTER 4 Concluions and Future work 4.1 Concluions 4.2 Future work The particles chain phenomenon in non-newtonian shear flow has been observed experimentally in [12] and [18]. The numerical model of three particles alignment in a viscoelastic fluid has been introduced in [9]. However, the numerical simulation of particle-chains in non-newtonian shear flow haven t been developed. I will use the 75

85 4.2. FUTURE WORK Oldroyd-B fluid with shear thinning to investigate the chain phenomenon of many particles with random initial positions. Figure 4.2.1: Particles alignment in polyisobutylene solution(j. Michele et al. Rheol. Acta 1977) Another extension of my current work will be the finitely extensible nonlinear elastic (FENE) dumbbell model in non-newtonian fluid flow. In order to study the properties of dilute polymer fluid, the motion of polymer molecules in the fluid is modeled as a suspension of dumbbells or spring chains with finite extensibility (e.g., see [1] and [2]). Besides the typical topics, such as the flow in the channel, flow in a cross-slot geometry and impacting drop problem, it would be interesting to consider the fluid-particle and particle-particle interactions in non-newtonian fluid flow of FENE type with particles. 76