Discontinuous shear thickening on dense non-brownian suspensions via lattice Boltzmann method

Transcription

1 Discontinuous shear thickening on dense non-brownian suspensions via lattice Boltzmann method Pradipto and Hisao Hayakawa Yukawa Insitute for Theoretical Physics Kyoto University Rheology of disordered particle part 1, YITP, June 22nd /38

2 Outline Introduction Outline of our model Hydrodynamics: lattice Boltzmann method Particle contacts Electrostatic repulsive forces Sheared suspensions simulations Results Discussions Future prospect 2/38

3 Suspensions 3/38

4 Suspensions Solid particles suspended in solvent fluid 3/38

5 Suspensions Solid particles suspended in solvent fluid Fluids are described by Stokes equation (Re 0): u = 0 p = η 2 u 3/38

6 Suspensions Solid particles suspended in solvent fluid Fluids are described by Stokes equation (Re 0): u = 0 p = η 2 u Suspensions rheology have unique behavior! 3/38

7 Discontinuous shear thickening 4/38

8 Discontinuous shear thickening (experimental observations) H. A. Barnes, J. Rheol. 33, 329 (1989). 5/38

9 Discontinuous shear thickening (experimental observations) R. G. Egres and N. J. Wagner J. Rheol. 49 3, (2015) H. A. Barnes, J. Rheol. 33, 329 (1989). 5/38

10 Discontinuous shear thickening (Stokesian Dynamics simulations) Seto. et al, Phys. Rev. Lett 111, (2013) Mari. et al, J. Rheol.58(6): (2014) 6/38

11 Discontinuous shear thickening (Stokesian Dynamics simulations) Seto. et al, Phys. Rev. Lett 111, (2013) Mari. et al, J. Rheol.58(6): (2014) 6/38

12 Discontinuous shear thickening (Stokesian Dynamics simulations) Seto. et al, Phys. Rev. Lett 111, (2013) Mari. et al, J. Rheol.58(6): (2014) 6/38

13 Discontinuous shear thickening 7/38

14 Discontinuous shear thickening No Theory 7/38

15 Discontinuous shear thickening No Theory Possible explanations: Lubrication (Brady s group) 7/38

16 Discontinuous shear thickening No Theory Possible explanations: Lubrication (Brady s group). Friction (R. Seto, et al, 2013) 7/38

17 Discontinuous shear thickening No Theory Possible explanations: Lubrication (Brady s group). Friction (R. Seto, et al, 2013) Granular systems (Otsuki and Hayakawa, 2011) 7/38

18 Discontinuous shear thickening No Theory Possible explanations: Lubrication (Brady s group). Friction (R. Seto, et al, 2013) Granular systems (Otsuki and Hayakawa, 2011) Brownian suspensions (R. Mari, et al, 2015) 7/38

19 Discontinuous shear thickening No Theory Possible explanations: Lubrication (Brady s group). Friction (R. Seto, et al, 2013) Granular systems (Otsuki and Hayakawa, 2011) Brownian suspensions (R. Mari, et al, 2015) Boundary effect (Brown and Jaeger, 2012) and (Allen, et al, 2017) (Brown s group) the contributions beneath the DST is still unclear Motivations: Recover DST using LBM 7/38

20 Discontinuous shear thickening No Theory Possible explanations: Lubrication (Brady s group). Friction (R. Seto, et al, 2013) Granular systems (Otsuki and Hayakawa, 2011) Brownian suspensions (R. Mari, et al, 2015) Boundary effect (Brown and Jaeger, 2012) and (Allen, et al, 2017) (Brown s group) the contributions beneath the DST is still unclear Motivations: Recover DST using LBM Decompose all contributions to the shear stress 7/38

21 Discontinuous shear thickening No Theory Possible explanations: Lubrication (Brady s group). Friction (R. Seto, et al, 2013) Granular systems (Otsuki and Hayakawa, 2011) Brownian suspensions (R. Mari, et al, 2015) Boundary effect (Brown and Jaeger, 2012) and (Allen, et al, 2017) (Brown s group) the contributions beneath the DST is still unclear Motivations: Recover DST using LBM Decompose all contributions to the shear stress Helpful for the theoretical construction 7/38

22 Lattice Boltzmann vs Stokesian Dynamics Lattice Boltzmann Stokesian Dynamics Particle motion obey the Newton equations. Hydrodynamic fields are calculated by solving the resistance matrix Separate lubrication calculation. 8/38

23 Lattice Boltzmann vs Stokesian Dynamics Lattice Boltzmann (susp3d) Stokesian Dynamics Particle motion obey the Newton equations. Hydrodynamic fields are calculated by solving the resistance matrix Separate lubrication calculation. Hydrodynamic fields are calculated locally at each lattice point. mesoscopic, possible to have simple local rules between fluid and solid Particle motion obey the Newton equations Separate lubrication calculation. 8/38

24 Outline of our model Equation of motions: m d ( ) U = ( ) Fα dt Ω T α α F α = F h + F c + F R α T α = T h + T c + T R α U(t + t) = U(t) + t m F(t) Ω(t + t) = Ω(t) + t m T(t) 9/38

25 Outline of our model Equation of motions: m d ( ) U = ( ) Fα dt Ω T α α F α = F h + F c + F R α Hydrodynamics (lattice Boltzmann) T α = T h + T c + T R α U(t + t) = U(t) + t m F(t) Ω(t + t) = Ω(t) + t m T(t) 9/38

26 Outline of our model Equation of motions: m d ( ) U = ( ) Fα dt Ω T α α F α = F h + F c + F R α Hydrodynamics (lattice Boltzmann) Contact between particles T α = T h + T c + T R α U(t + t) = U(t) + t m F(t) Ω(t + t) = Ω(t) + t m T(t) 9/38

27 Outline of our model Equation of motions: m d ( ) U = ( ) Fα dt Ω T α α F α = F h + F c + F R α Hydrodynamics (lattice Boltzmann) Contact between particles T α = T h + T c + T R α U(t + t) = U(t) + t m F(t) Ω(t + t) = Ω(t) + t m T(t) Electrostatic repulsive forces 9/38

28 Outline of our model Equation of motions: m d ( ) U = ( ) Fα dt Ω T α α F α = F h + F c + F R α Hydrodynamics (lattice Boltzmann) Contact between particles T α = T h + T c + T R α U(t + t) = U(t) + t m F(t) Ω(t + t) = Ω(t) + t m T(t) Electrostatic repulsive forces 9/38

29 Hydrodynamics: lattice Boltzmann method Hydrodynamics (lattice Boltzmann) 10/38

30 Hydrodynamics: lattice Boltzmann method Hydrodynamics (lattice Boltzmann) By using LBM, we need to discretize the unit length a (particle radius) to lattice unit x! 10/38

31 Hydrodynamics: lattice Boltzmann method Hydrodynamics (lattice Boltzmann) By using LBM, we need to discretize the unit length a (particle radius) to lattice unit x! Our simulation used x = 0.1a 10/38

32 Hydrodynamics: lattice Boltzmann method Hydrodynamics (lattice Boltzmann) By using LBM, we need to discretize the unit length a (particle radius) to lattice unit x! Our simulation used x = 0.1a All quantities below are written in lattice units! 10/38

33 History of LBM Initially devised as an extension of the Lattice Gas Automata by McNamara and Zanetti (McNamara, Zanetti, 1988) 11/38

34 History of LBM Initially devised as an extension of the Lattice Gas Automata by McNamara and Zanetti (McNamara, Zanetti, 1988) Now, LBM is widely used for various computational fluid dynamics simulation. 11/38

35 The discrete distribution function (He and Luo, 1997) n particle velocity (v) distribution function on space time point (r, t) 12/38

36 The discrete distribution function (He and Luo, 1997) n particle velocity (v) distribution function on space time point (r, t) time evolution of n in continuous space: ( ) dn t n + v n = dt coll 12/38

37 The discrete distribution function (He and Luo, 1997) n particle velocity (v) distribution function on space time point (r, t) time evolution of n in continuous space: ( ) dn t n + v n = dt coll we can project to the Hermite bases: n(r, v, t) = ω(v) l=0 1 l! A(l) H (l) (v) 12/38

38 The discrete distribution function (He and Luo, 1997) n particle velocity (v) distribution function on space time point (r, t) time evolution of n in continuous space: ( ) dn t n + v n = dt coll we can project to the Hermite bases: n(r, v, t) = ω(v) l=0 1 l! A(l) H (l) (v) the expansion coefficients: A (l) (r, t) = dvn(r, v, t)h (l) (v) 12/38

39 The discrete distribution function (He and Luo, 1997) n particle velocity (v) distribution function on space time point (r, t) time evolution of n in continuous space: ( ) dn t n + v n = dt coll we can project to the Hermite bases: n(r, v, t) = ω(v) l=0 1 l! A(l) H (l) (v) the expansion coefficients: A (l) (r, t) = dvn(r, v, t)h (l) (v) which is linear combination of the moments of n The first few of the Hermite polynomials (H (l) (v)): H (0) (v) = 1 H (1) (v) = v α H (2) (v) = v α v β c 2 s δ αβ the Gaussian weight function: ( ) 1 ω(v) = exp v2 (2π) D/2 2 12/38

40 The discrete distribution function (He and Luo, 1997) n particle velocity (v) distribution function on space time point (r, t) time evolution of n in continuous space: ( ) dn t n + v n = dt coll we can project to the Hermite bases: n(r, v, t) = ω(v) l=0 1 l! A(l) H (l) (v) the expansion coefficients: A (l) (r, t) = dvn(r, v, t)h (l) (v) which is linear combination of the moments of n The first few of the Hermite polynomials (H (l) (v)): H (0) (v) = 1 H (1) (v) = v α H (2) (v) = v α v β c 2 s δ αβ the Gaussian weight function: ( ) 1 ω(v) = exp v2 (2π) D/2 2 12/38

41 From Boltzmann equation to LBE Truncate: (to several orders of K) n(r, v, t) = ω(v) K l=0 1 l! A(l) H (l) (v) 13/38

42 From Boltzmann equation to LBE Truncate: (to several orders of K) n(r, v, t) = ω(v) K l=0 only finite sets of velocities are needed! 1 l! A(l) H (l) (v) 13/38

43 From Boltzmann equation to LBE Truncate: (to several orders of K) n(r, v, t) = ω(v) K l=0 1 l! A(l) H (l) (v) only finite sets of velocities are needed! we can compute the integral for A with Gauss-Hermite quadrature. (p(v) arbitrary polynomial of v, c i discrete lattice velocity, w i quadrature weight ) dvω(v)p(v) = b w i p(c i ) i=1 13/38

44 From Boltzmann equation to LBE Truncate: (to several orders of K) n(r, v, t) = ω(v) K l=0 1 l! A(l) H (l) (v) only finite sets of velocities are needed! we can compute the integral for A with Gauss-Hermite quadrature. (p(v) arbitrary polynomial of v, c i discrete lattice velocity, w i quadrature weight ) dvω(v)p(v) = b w i p(c i ) i=1 dvn(r, v, t)v v = dv ω(v) n(r, v, t)v v ω(v) b n(r, v, t) = w i c i c i ω(v) = i=1 b n i (r, c i, t)c i c i i=1 13/38

45 From Boltzmann equation to LBE Truncate: (to several orders of K) n(r, v, t) = ω(v) K l=0 1 l! A(l) H (l) (v) only finite sets of velocities are needed! we can compute the integral for A with Gauss-Hermite quadrature. (p(v) arbitrary polynomial of v, c i discrete lattice velocity, w i quadrature weight ) dvω(v)p(v) = b w i p(c i ) i=1 dvn(r, v, t)v v = dv ω(v) n(r, v, t)v v ω(v) b n(r, v, t) = w i c i c i ω(v) = i=1 b n i (r, c i, t)c i c i discrete distribution function n i (r, c i, t) = w i n(r, v, t)/ω(v) i=1 13/38

46 the lattice Boltzmann method Evolving equation: n i (r + c i, t + 1) = n i (r, t) + i (r, t) 14/38

47 the lattice Boltzmann method Evolving equation: n i (r + c i, t + 1) = n i (r, t) + i (r, t) streaming 14/38

48 the lattice Boltzmann method Evolving equation: n i (r + c i, t + 1) = n i (r, t) + i (r, t) streaming collision 14/38

49 the lattice Boltzmann method Evolving equation: n i (r + c i, t + 1) = n i (r, t) + i (r, t) streaming collision Hydrodynamic fields: mass density ρ = i n i momentum density j = i n i c i momentum flux Π = i n i c i c i 14/38

50 LBM: Collision operator Linearized collision operator i (n) = i (n eq i ) + j L ij n neq j 15/38

51 LBM: Collision operator Linearized collision operator i (n) = i (n eq i ) + j L ij n neq j First, we find the discrete equilibrium distribution function n eq i, 15/38

52 LBM: Collision operator Linearized collision operator i (n) = i (n eq i ) + j L ij n neq j First, we find the discrete equilibrium distribution function n eq i, Maxwell-Boltzmann distribution: ( ) 3 ( ) n eq m 2 m(v u) (v u) = ρ exp 2πk b T 2k b T 15/38

53 LBM: Collision operator Linearized collision operator i (n) = i (n eq i ) + j L ij n neq j First, we find the discrete equilibrium distribution function n eq i, Maxwell-Boltzmann distribution: ( ) 3 ( ) n eq m 2 m(v u) (v u) = ρ exp 2πk b T 2k b T Expanding with Hermite polynomials up to 2nd order, with coefficients: A 0 eq = ρ A 1 eq = ρu A 2 eq = ρ(u c 2 s ) 15/38

54 LBM: Collision operator Linearized collision operator i (n) = i (n eq i ) + j L ij n neq j First, we find the discrete equilibrium distribution function n eq i, Maxwell-Boltzmann distribution: ( ) 3 ( ) n eq m 2 m(v u) (v u) = ρ exp 2πk b T 2k b T Expanding with Hermite polynomials up to 2nd order, with coefficients: A 0 eq = ρ A 1 eq = ρu A 2 eq = ρ(u c 2 s ) discrete equilibrium distribution function: ( ) n eq i = a c i ρ+ j c i + (ρuu) : (c ic i cs 2 1) cs 2 2cs 4 15/38

55 LBM: Collision operator Linearized collision operator i (n) = i (n eq i ) + j L ij n neq j First, we find the discrete equilibrium distribution function n eq i, Maxwell-Boltzmann distribution: ( ) 3 ( ) n eq m 2 m(v u) (v u) = ρ exp 2πk b T 2k b T Expanding with Hermite polynomials up to 2nd order, with coefficients: A 0 eq = ρ A 1 eq = ρu A 2 eq = ρ(u c 2 s ) discrete equilibrium distribution function: ( ) n eq i = a c i ρ+ j c i + (ρuu) : (c ic i cs 2 1) cs 2 2cs 4 Weight coefficients: a 0 = 12 a 1 = 2 a 2 = 1 c s = k b T lattice sound speed A : B = Tr(AB) 15/38

56 LBM: Collision operator Linearized collision operator i (n) = i (n eq ) + j L ij n neq j 16/38

57 LBM: Collision operator Linearized collision operator i (n) = i (n eq ) + j L ij n neq j not necessary to construct and calculate L ij, 16/38

58 LBM: Collision operator Linearized collision operator i (n) = i (n eq ) + j L ij n neq j not necessary to construct and calculate L ij, use its eigen equation instead! L ij = 0 c i L ij = 0 c i c i L ij = λc j c j ci 2 L ij = λ ν cj 2 i i i i c j c j traceless 16/38

59 LBM: Collision operator Linearized collision operator i (n) = i (n eq ) + j L ij n neq j not necessary to construct and calculate L ij, use its eigen equation instead! L ij = 0 c i L ij = 0 c i c i L ij = λc j c j ci 2 L ij = λ ν cj 2 i c j c j traceless i discrete equilibrium distribution function: ( n eq i = a c i ρ + j c i c 2 s i + (ρuu) : (c ic i c 2 s 1) 2c 4 s ) i 16/38

60 LBM: Collision operator Linearized collision operator i (n) = i (n eq ) + j L ij n neq j not necessary to construct and calculate L ij, use its eigen equation instead! L ij = 0 c i L ij = 0 c i c i L ij = λc j c j ci 2 L ij = λ ν cj 2 i c j c j traceless i discrete equilibrium distribution function: ( n eq i = a c i Only the second moment of n eq i ρ + j c i c 2 s i + (ρuu) : (c ic i c 2 s 1) 2c 4 s that is affected by collision ) i 16/38

61 LBM: Collision operator Linearized collision operator i (n) = i (n eq ) + j L ij n neq j not necessary to construct and calculate L ij, use its eigen equation instead! L ij = 0 c i L ij = 0 c i c i L ij = λc j c j ci 2 L ij = λ ν cj 2 i c j c j traceless i discrete equilibrium distribution function: ( n eq i = a c i ρ + j c i c 2 s i + (ρuu) : (c ic i c 2 s 1) 2c 4 s Only the second moment of n eq i that is affected by collision Post collision distribution function: n i = a c i ( ρ + j c i c 2 s ) + (ρuu + Πneq, ) : (c i c i c 2 s 1) 2c 4 s i ) 16/38

62 LBM: Collision operator Linearized collision operator i (n) = i (n eq ) + j L ij n neq j not necessary to construct and calculate L ij, use its eigen equation instead! L ij = 0 c i L ij = 0 c i c i L ij = λc j c j ci 2 L ij = λ ν cj 2 i i i i c j c j traceless discrete equilibrium distribution function: ( n eq i = a c i ρ + j c i c 2 s + (ρuu) : (c ic i c 2 s 1) 2c 4 s Only the second moment of n eq i that is affected by collision Post collision distribution function: n i = a c i ( ρ + j c i c 2 s ) + (ρuu + Πneq, ) : (c i c i c 2 s 1) 2c 4 s Calculate Π neq, ) 16/38

63 LBM: Collision operator Π neq, Π neq = Π Π eq Π = i n ic i c i 17/38

64 LBM: Collision operator Π neq, Π neq = Π Π eq Π = i n ic i c i nonequilibrium second moments obtained from the eigen equation of L ij : L ij = 0 c i L ij = 0 c i c i L ij = λc j c j ci 2 L ij = λ ν cj 2 i i i i 17/38

65 LBM: Collision operator Π neq, Π neq = Π Π eq Π = i n ic i c i nonequilibrium second moments obtained from the eigen equation of L ij : L ij = 0 c i L ij = 0 c i c i L ij = λc j c j ci 2 L ij = λ ν cj 2 i i i i Π neq, = (1 + λ) Π neq (1 + λ ν)(π neq : 1)1 17/38

66 LBM: Collision operator Π neq, Π neq = Π Π eq Π = i n ic i c i nonequilibrium second moments obtained from the eigen equation of L ij : L ij = 0 i c i L ij = 0 i c i c i L ij = λc j c j i Π neq, = (1 + λ) Π neq (1 + λ ν)(π neq : 1)1 λ and λ ν are related to shear η and bulk η ν viscosities from the multiscale analysis: ( 1 η = ρcs 2 t λ + 1 ) ( 2 η ν = ρcs 2 t + 1 ) 2 3λ ν 3 ci 2 L ij = λ ν cj 2 i 17/38

67 LBM: Collision operator Π neq, Π neq = Π Π eq Π = i n ic i c i nonequilibrium second moments obtained from the eigen equation of L ij : L ij = 0 i c i L ij = 0 i c i c i L ij = λc j c j i Π neq, = (1 + λ) Π neq (1 + λ ν)(π neq : 1)1 λ and λ ν are related to shear η and bulk η ν viscosities from the multiscale analysis: ( 1 η = ρcs 2 t λ + 1 ) ( 2 η ν = ρcs 2 t + 1 ) 2 3λ ν 3 What will happen on the solid boundary conditions (surface of the particles)? ci 2 L ij = λ ν cj 2 i 17/38

68 Solid-fluid boundary condition Anthony Ladd s bounce-back rule: n b (r, t + t) = n b (r, t) 2ac bρ 0 u b c b c 2 s Velocity of the boundary nodes: u b = U + Ω (r b R) R is the center of mass of the particle 18/38

69 Solid-fluid boundary condition Forces exerted at the boundary nodes: [ ] f(r b, t+ 1 3 x t) = 2nb (r, t) 2ac b ρ 0u b c b 2 t cs 2 c b 19/38

70 Solid-fluid boundary condition Sum over all boundary nodes within a particle: Forces exerted at the boundary nodes: [ ] f(r b, t+ 1 3 x t) = 2nb (r, t) 2ac b ρ 0u b c b 2 t cs 2 c b 19/38

71 Solid-fluid boundary condition Sum over all boundary nodes within a particle: Forces exerted at the boundary nodes: [ ] f(r b, t+ 1 3 x t) = 2nb (r, t) 2ac b ρ 0u b c b 2 t cs 2 c b F h = b T h = b σ h = b f(r b ) r b f(r b ) r b f(r b ) 19/38

72 Shared nodes 20/38

73 Shared nodes need separated lubrication forces calculation! 20/38

74 Lubrications If gap between particle is less than 1 lattice unit.. 21/38

75 Lubrications If gap between particle is less than 1 lattice unit.. Grand-resistance formulation (Nguyen and Ladd, 2002) 21/38

76 Lubrications If gap between particle is less than 1 lattice unit.. Grand-resistance formulation (Nguyen and Ladd, 2002) F 1 A 11 B 11 B 22 T 1 B 11 C 11 C 12 T 2 S 1 S 2 = B 22 C 12 C 22 G 11 H 11 H 12 G 22 H 21 H 22 (Kim and Karilla, 1991) U 12 Ω 1 Ω 2 21/38

77 Lubrications If gap between particle is less than 1 lattice unit.. Grand-resistance formulation (Nguyen and Ladd, 2002) F 1 A 11 B 11 B 22 T 1 B 11 C 11 C 12 T 2 S 1 S 2 U 12 = U 1 U 2 relative velocity = B 22 C 12 C 22 G 11 H 11 H 12 G 22 H 21 H 22 (Kim and Karilla, 1991) U 12 Ω 1 Ω 2 21/38

78 Lubrications If gap between particle is less than 1 lattice unit.. Grand-resistance formulation (Nguyen and Ladd, 2002) F 1 A 11 B 11 B 22 T 1 B 11 C 11 C 12 T 2 S 1 S 2 U 12 = U 1 U 2 relative velocity = B 22 C 12 C 22 G 11 H 11 H 12 G 22 H 21 H 22 (Kim and Karilla, 1991) U 12 Ω 1 Ω 2 Each coefficients can be expressed in terms of scalar function i.e 21/38

79 Lubrications If gap between particle is less than 1 lattice unit.. Grand-resistance formulation (Nguyen and Ladd, 2002) F 1 A 11 B 11 B 22 T 1 B 11 C 11 C 12 T 2 S 1 S 2 U 12 = U 1 U 2 relative velocity = B 22 C 12 C 22 G 11 H 11 H 12 G 22 H 21 H 22 (Kim and Karilla, 1991) U 12 Ω 1 Ω 2 Each coefficients can be expressed in terms of scalar function i.e H 12 = Y H 12(ɛ αγδ d δ d β + ɛ βγδ d δ d α ) d displacement unit vector along axis. ɛ Levi-Civita symbol 21/38

80 Lubrications If gap between particle is less than 1 lattice unit.. Grand-resistance formulation (Nguyen and Ladd, 2002) F 1 A 11 B 11 B 22 T 1 B 11 C 11 C 12 T 2 S 1 S 2 U 12 = U 1 U 2 relative velocity = B 22 C 12 C 22 G 11 H 11 H 12 G 22 H 21 H 22 (Kim and Karilla, 1991) U 12 Ω 1 Ω 2 Each coefficients can be expressed in terms of scalar function i.e H 12 = Y H 12(ɛ αγδ d δ d β + ɛ βγδ d δ d α ) d displacement unit vector along axis. ɛ Levi-Civita symbol Each scalar function is a function of gap h and β = a i a j ( ) 1 Y12 H 2β 2 (1 + 7β) = 8 log πηa i h 5(1 + β) 5 i.e 21/38

81 Lubrications Cutoff length δ = acontact a hydro a contact to allow contact 22/38

82 Lubrications Cutoff length δ = acontact a hydro a contact to allow contact We used δ = /38

83 Particle contacts 23/38

84 Contact model Linear spring dashpot model (Luding, 2008) 24/38

85 Contact model Linear spring dashpot model (Luding, 2008) F c ij = Fnor ij + F tan ij (Fleischmann, 2015) 24/38

86 Contact model Coulomb friction rules: F tan ij µ( F nor ij ) 25/38

87 Contact model Coulomb friction rules: F tan ij µ( F nor ij ) slip 25/38

88 Contact model Coulomb friction rules: F tan ij µ( F nor ij ) slip F tan ij µ( F nor ij ) 25/38

89 Contact model Coulomb friction rules: F tan ij µ( F nor ij ) slip F tan ij µ( F nor ij ) stick 25/38

90 Contact model Coulomb friction rules: F tan ij µ( F nor ij ) slip F tan ij µ( F nor ij ) stick Stress contribution from contact: Normal: σ nor αβ = 1 2V i j i (r ij,α F nor ij,β + r ij,βf nor ij,α) 25/38

91 Contact model Coulomb friction rules: F tan ij µ( F nor ij ) slip F tan ij µ( F nor ij ) stick Stress contribution from contact: Normal: σ nor αβ = 1 2V i j i (r ij,α F nor ij,β + r ij,βf nor ij,α) Tangential: σ tan αβ = 1 V i j i r ij,α F tan ij,β 25/38

92 Contact model Coulomb friction rules: F tan ij µ( F nor ij ) slip F tan ij µ( F nor ij ) stick Stress contribution from contact: Normal: σ nor αβ = 1 2V i j i (r ij,α F nor ij,β + r ij,βf nor ij,α) Tangential: σ tan αβ = 1 V i j i r ij,α F tan ij,β 25/38

93 Electrostatic repulsive forces (Israelachvili, 2001) 26/38

94 Electrostatic repulsive forces (Israelachvili, 2001) ( ) F R ij = 1 a i a j λ a i + a j F exp( h/λ)ˆn ij λ Debye length 26/38

95 Electrostatic repulsive forces (Israelachvili, 2001) ( ) F R ij = 1 a i a j λ a i + a j F exp( h/λ)ˆn ij λ Debye length we used λ = 0.2a 26/38

96 Electrostatic repulsive forces (Israelachvili, 2001) ( ) F R ij = 1 a i a j λ a i + a j F exp( h/λ)ˆn ij λ Debye length we used λ = 0.2a Stress contribution: σ R αβ = 1 V R ij,α F R ij,β i j i 26/38

97 Sheared suspensions 27/38

98 Sheared suspensions wall moves with velocity u wall to x and x directions All simulation use N=512 particles. 27/38

99 Rheology Shear stress: σ αβ = σ h αβ + σtan αβ + σnor αβ + σr αβ 28/38

100 Rheology Shear stress: σ αβ = σ h αβ + σtan αβ + σnor αβ + σr αβ Apparent viscosity: η = σ αβ / γ 28/38

101 Rheology Shear stress: σ αβ = σ h αβ + σtan αβ + σnor αβ + σr αβ Apparent viscosity: η = σ αβ / γ Dimensionless shear rate: γ = 6πη 0a 2 γ F 28/38

102 Rheology Shear stress: σ αβ = σ h αβ + σtan αβ + σnor αβ + σr αβ Apparent viscosity: η = σ αβ / γ Dimensionless shear rate: γ = 6πη 0a 2 γ F 28/38

103 Sheared Suspensions: (Mari, et al, 2014) η η 0 vs γ, compared with results from 29/38

104 Sheared Suspensions: (Mari, et al, 2014) η η 0 vs γ, compared with results from 29/38

105 η η 0 Sheared Suspensions: vs σ/η 0 γ, compared with results from (Mari, et al, 2014) 30/38

106 η η 0 Sheared Suspensions: vs σ/η 0 γ, compared with results from (Mari, et al, 2014) 30/38

107 Contributions to shear viscosity for φ = /38

108 Contributions to shear viscosity for φ = /38

109 Contributions to shear viscosity for φ = 0.54 and /38

110 Contributions to shear viscosity for φ = 0.54 and /38

111 Time evolution of shear viscosity φ = 0.57 and φ = /38

112 Time evolution of shear viscosity φ = 0.57 and φ = /38

113 Time evolution of contact number for φ = /38

114 Time evolution of contact number for φ = /38

115 Evolution of the contact network (φ = 0.57) high shear rate low shear rate 35/38

116 Discussions Slight deviation from Seto s 36/38

117 Discussions Slight deviation from Seto s Need to allow separate switching length between boundary nodes and lubrication calculation for normal, tangential, and rotation modes 36/38

118 Discussions Slight deviation from Seto s Need to allow separate switching length between boundary nodes and lubrication calculation for normal, tangential, and rotation modes We need to find a suitable contact spring constant for every φ 36/38

119 Discussions Slight deviation from Seto s Need to allow separate switching length between boundary nodes and lubrication calculation for normal, tangential, and rotation modes We need to find a suitable contact spring constant for every φ Contact force dominates the shear thickening regime 36/38

120 Discussions Slight deviation from Seto s Need to allow separate switching length between boundary nodes and lubrication calculation for normal, tangential, and rotation modes We need to find a suitable contact spring constant for every φ Contact force dominates the shear thickening regime Number of contact increased on high shear rate, creates force chains from one side to another. 36/38

121 Discussions Slight deviation from Seto s Need to allow separate switching length between boundary nodes and lubrication calculation for normal, tangential, and rotation modes We need to find a suitable contact spring constant for every φ Contact force dominates the shear thickening regime Number of contact increased on high shear rate, creates force chains from one side to another. Electsrotatic repulsive force shear thinning at low shear rate 36/38

122 Discussions Slight deviation from Seto s Need to allow separate switching length between boundary nodes and lubrication calculation for normal, tangential, and rotation modes We need to find a suitable contact spring constant for every φ Contact force dominates the shear thickening regime Number of contact increased on high shear rate, creates force chains from one side to another. Electsrotatic repulsive force shear thinning at low shear rate The computational cost highly depends on the size. 36/38

123 Discussions Slight deviation from Seto s Need to allow separate switching length between boundary nodes and lubrication calculation for normal, tangential, and rotation modes We need to find a suitable contact spring constant for every φ Contact force dominates the shear thickening regime Number of contact increased on high shear rate, creates force chains from one side to another. Electsrotatic repulsive force shear thinning at low shear rate The computational cost highly depends on the size. Our simulation used x = 0.1a, to let Re < 1, one need typical particle speed U < x t. 36/38

124 Discussions Slight deviation from Seto s Need to allow separate switching length between boundary nodes and lubrication calculation for normal, tangential, and rotation modes We need to find a suitable contact spring constant for every φ Contact force dominates the shear thickening regime Number of contact increased on high shear rate, creates force chains from one side to another. Electsrotatic repulsive force shear thinning at low shear rate The computational cost highly depends on the size. Our simulation used x = 0.1a, to let Re < 1, one need typical particle speed U < x t. To travel a distance of its radius, one particles needs > 6200 time steps 36/38

125 Future prospects Good starting point for the theoretical works 37/38

126 Future prospects Good starting point for the theoretical works friction scenario. 37/38

127 Future prospects Good starting point for the theoretical works friction scenario. Analyze the percolation of the contact network 37/38

128 Future prospects Good starting point for the theoretical works friction scenario. Analyze the percolation of the contact network Implement contact with rolling friction. 37/38

129 Future prospects Good starting point for the theoretical works friction scenario. Analyze the percolation of the contact network Implement contact with rolling friction. Use more realistic boundary conditions 37/38

130 References Ladd, A.J.C Numerical simulations of particulate suspensions via a discretized Boltzmann equation. Part 1. Theoretical foundation, Journal of fluid mechanics, vol 271, , Ladd, A.J.C Numerical simulations of particulate suspensions via a discretized Boltzmann equation. Part 2. Numerical Results, Journal of fluid mechanics, vol 271, , N.Q. Nguyen and Ladd, A.J.C, Lubrication corrections for lattice-boltzmann simulations of particle suspensions. Physical Review E 66, , Ladd, A.J.C, Susp3D, an open source code for simulating suspensions based on Lattice Boltzmann Method, Romain Mari, Ryohei Seto, Jeffrey F. Morris and Morton M. Denn, Shear thickening, frictionless and frictional rheologies in non-brownian suspensions, Journal of Rheology 58(6), , Stefan Luding, Cohesive, frictional powders: contact models for tension, Granular Matter 10:235246, Ryohei Seto,Romain Mari,Jeffrey F. Morris, and Morton M. Denn, Discontinuous Shear Thickening of Frictional Hard-Sphere Suspensions, Physical Review Letters 111, , S. Kim and S. J. Karrila, Microhydrodynamics: Principles and Selected Applications, Butterworth-Heinemann, Boston, MA, /38