Microstructure and Particle-Phase Stress in a Dense Suspension

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1 Microstructure and Particle-Phase Stress in a Dense Suspension Jim Jenkins, Cornell University Luigi La Ragione, Politecnico di Bari Predict microstructure and particle stresses in slow, steady, pure straining of a dense, single layer of spheres in a viscous fluid Singh & Nott, J Fluid Mech. 412, 279 (2) Jenkins & La Ragione, J. Fluid Mech. 763, 218 (215)

2 Pure Straining θ B d (BA) A BA) Area fraction: ν Average number of near contacts: k Short-range repulsive force: F

3 Elements of the Analysis Pair translations and rotations Pair forces and moments Force and moment balances Averages over directions and neighbors for equilibrium Rate equations for the separation and orientation Distribution of particle orientation and particle flux Averaging over particle orientations for the stress

4 Kinematics Treat BA exactly Relative velocity of centers of pair BA v α (BA) = d dt s(ba) ˆdα (BA) + 2a d dt θ(ba)ˆt α (BA) Relative velocity of points of near contact of BA v (BA) α + a( ω (A) + ω (B) (BA) )ˆt α Treat na on average Relative velocity of centers of pair na v i (na) = 2aD αβ ˆdβ (na) Relative velocity of points of near contact of na v α (na) + aω (A)ˆt α (na)

5 Interaction Forces Jeffrey & Onishi (1984), Jeffrey (1992) Pair BA F (BA) α = 6πµaK (BA) αβ v (BA) β F ˆd (BA) s (BA) α 9.54πµa 2 ( ˆt (BA) (BA) (BA) β D βγ ˆd γ )ˆt α +πµa 2 ln K (BA) = 1 αβ 4 a s (BA).96 ω(a)ˆt (BA) α + πµa 2 ln a ˆd (BA) s (BA) α ˆd (BA) β ln a s (BA) a s (BA) ˆt (BA) α (BA)ˆt β ω(b)ˆt (BA) α Pairs na F (na) α = 12πµa 2 K (na) αβ D βγ ˆd (na) γ + πµa 2 F s ˆd α (na) + 2πµa 2 K (na) αβ = 1 a 4 ln a s.96 ω(a)ˆt α (na) ln 1 s.96 ( ˆt (na) (na) (na) β D βγ ˆd γ )ˆt α s ˆd (na) α ˆd (na) β ln a s ˆt (na) α (na)ˆt β

6 Force Balance, Particle A N ( A ) n B F α (BA) + F α (na) = F α (AB) = F α (BA), ˆd α (AB) = ˆd α (BA), ˆt α (AB) = ˆt α (BA) = 4πµa 2 ln +πµa 2 Difference of force balances = 3πµa a s (BA) 2ln +6πµa 2 a s a s (BA) Normal component d dt s(ba) 2 F s (BA) ˆd (BA) α J αβγ D βγ 2 F s Y (BA) ˆd α α Tangential component a s (BA) d dt θ(ba) 19πµa 2ˆt (BA) (BA) α D ˆdβ αβ.96 S + 6πµa a 2 s ˆt (BA) α J αβγ D βγ (S = ω A + ω B )

7 Moment Balance, Particle A N ( A ) n B ε αβ F α (BA) ˆdβ (BA) + ε αβ F α (na) ˆdβ (na) = (ε 12 = - ε 12 = 1, ε 11 = ε 22 = ) = 2 ln + ln a s (BA) a s (BA) Sum of Moment Balances d dt θ(ba) 9.54ˆt (BA) (BA) α D ˆdβ αβ + k 1 2 ln a s S + 2 ln a s.96 ε A D βγ αβ αγ

8 Neighborhood Averages Isotropic representations Y α N ( A ) n B ˆdα (na) = ξˆd α (BA) A αβ N ( A ) n B ˆdα (na) ˆd β (na) = β 1 δ αβ + β 2 ˆdα (BA) ˆdβ (BA) J αβγ N ( A ) n B ˆdα (na) ˆdβ (na) ˆd γ (na) (BA) (BA) = α ˆdβ 1ˆdα ˆd (BA) γ + α ( 2 δ ˆdk (BA) αβ + δ ˆdβ (BA) αγ + δ ) (BA) ˆdα βγ

9 Neighborhood Averages s a = π 8νg g = 16 7ν 16(1 ν) 2 Isotropic distribution function π Ψ α..γ = 2 I( θ )ˆd α..ˆd γ d θ I( θ ) =, < θ < π / 3 3(k 1) / 4π, π / 3 < θ < π α 1 =, α 2 = 3 3(k 1) 16π α β 1 = (k 1) (k 1) 16π, β 2 = 3 3(k 1) 8π = 2α ξ = 3 3(k 1) 4π = 4α

10 Trajectory Equations!γ D αβ =!γ ˆd (BA) α D ˆdβ (BA) αβ =!γ cos2θ, ˆd α (BA) = (sinθ,cosθ) ˆt α (BA) D αβ ˆdβ (BA) =!γ sin2θ Lengths dimensionless by a and F F / ( πµa 3!γ ) 1 s ds dγ = 2 3 F 1 s + 4 α s + 4 α s cos2θ ln 1 s dθ dγ = α 1 s sin2θ 1 2 ln 1 s S Tangential force difference - Sum moments S = 12α (4α k +1) 1/ s ln(1/ s).96 sin2θ

11 Contact Distribution A(θ)dθ: number of contacts within dθ A(θ) dθ dγ = constant and A(θ) π/2 dθ = k 4 Knowledge of k determines the constant of integration. Implement as differential equations I(θ) A( θ ) d θ θ di dθ = A, I() =, I(π / 2) = k / 4 da dθ = A d θ! θ! dθ d θ! dθ =! θ θ +! θ s ds dθ

12 Differential Equations ds dθ = ds / dγ dθ / dγ f[s,θ; s(ν),k] da dθ = A dθ / dγ θ dθ dγ + s dθ dγ ds dθ di dθ = A dγ dθ = 1 dθ / dγ Boundary Conditions x 1 θ = θ + (θ 1 θ )x s(θ ) =.3 s(θ 1 ) =.3 I(θ ) = I(θ 1 ) = k / 4 γ(θ ) = γ(θ 1 ) = 1.1

13 Separation \ k = 3., γ = 1.1, F = 1-4 ν =.55 ν =.6 ν =.65 2 s θ The repulsive force creates asymmetry about π/4.

14 Inverse Separation k = 3., γ = 1.1, F = 1-4 ν =.55 ν =.6 ν = /s θ The asymmetry of the separation clarified.

15 Angular Velocity 5 k = 3., γ = 1.1, F = dθ/dγ ν =.55 ν =.6 ν = θ A(θ) dθ dγ = constant

16 Circumferential Distribution 4 35 k = 3., γ = 1.1, s = s 1 =.3 A, F = A+1, F = θ The repulsive force shifts the anisotropy due to the flow, symmetric about π/4, to a distribution that is asymmetric.

17 Circumferential Distribution k = 3., γ = 1.1, F = 1-4 ν =.55 ν =.6 ν =.65 A θ Asymmetry about π/4 in both the distribution and interval.

18 T αβ = na Stress (dimensional) 2π A(θ)F α (BA) ˆdβ (BA) dθ n: number/area F α (BA) = N ( A ) (na) F α n B t αβ = T αβ aµ!γ, n = ν πa 2, F = F πa 3 µ!γ t αβ = ν α 2π A(θ) ( 4F + 6cos2θ) s ˆd ˆdβ α dθ +ν s α 2π A(θ)sin2θ ˆt ˆdβ α dθ +ν 12α (3α k +1) 1/ s ln(1/ s) 2π A(θ)ln 1 s sin2θ ˆt ˆdβ α dθ ˆd α (AB) = (sinθ,cosθ), ˆt α (AB) = (cosθ, sinθ)

19 Pressure p 1 2 (t xx + t yy ) p = 4ν α s π/2 ( ) A(θ) 2F + 3cos2θ dθ Shear stress τ 1 2 (t xx t yy ) π/2 ( ) τ = 4ν α A(θ) 2F + 3cos2θ cos2θdθ s +ν α 2π s A(θ)sin 2 2θdθ +ν 6α (3α k +1) 1/ s ln(1/ s) 2π A(θ)ln 1 s sin2 2θdθ

20 Stresses 25 k = 3., γ = 1.1, F = 1-4 Pressure (o), Shear Stress (x) ν Asymmetry about π/4 creates particle pressure.

21 Initial/Final Separation 2 k = 3., γ = 1.1, F = 1-4 Pressure (o), Shear Stress (x) s, s 1 The initial/final separation of the typical trajectory is where the shear stress exhibits a maximum (.3).

22 Conclusions A rough implementation of particle equilibrium for particles assumed to interact as pairs through points of near contact produces expressions for the normal and tangential components of their relative velocities as functions of their orientation with respect to the flow. From these velocities, the separation of neighboring particles and their circumferential distribution can be determined as functions of orientation. The existence of asymmetries in the circumferential distribution of particles influences the shear stress and produces a pressure associated with viscous interactions between particle pairs at points of near contact.

Chapter 6: Momentum Analysis

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