Closure relations for non-uniform suspensions Kengo Ichiki Andrea - PowerPoint PPT Presentation

Closure relations for non-uniform suspensions Kengo Ichiki Andrea Prosperetti Department of Mechanical Engineering The Johns Hopkins University Supported by DOE grant FG02-99ER14966 November 25, 2003 at APS / DFD (New Jersey)

Introduction Non-uniform suspensions . Practically important: • Shear-induced di ff usivity • Particle migration in Stokes flows • Stratification in sedimentation Uniform suspension is too simple: ^ k • No strain in sedimentation • No slip velocity in shear problem φ φ Important physics vanishes! 0 November 25, 2003 APS / DFD at New Jersey Page 2

Introduction Goal: To derive the constitutive equations of S : viscous stress of the mixture F : interphase force valid for all sedimentation , torque , and shear problems from first-principle simulations by Stokesian Dynamics method (Mo-Sangani 1994) under periodic boundary condition for random hard-sphere configurations with non-uniform weight References: Marchioro et al. , Int. J. Multiphase Flow 26 (2000) 783; 27 (2001) 237. Ichiki and Prosperetti, submitted to Phys. Fluids . November 25, 2003 APS / DFD at New Jersey Page 3

Rheology Uniform suspensions — Shear S µ = 2 µ e E m 1 � ∇ u m + ( ∇ u m ) † � E m = 2 u m : mixture velocity µ : viscosity of the fluid µ e : relative viscosity of the mixture Non-uniform suspensions – Sedimentation E m � 0 and µ e plays a role November 25, 2003 APS / DFD at New Jersey Page 4

Viscous Stress S Closure relation 1 S � ∇ u ∆ + ( ∇ u ∆ ) † � E ∆ = = 2 µ e E m 2 µ − 1 3 ( ∇ · u ∆ ) I + 2 µ ∆ E ∆ 1 � u ∆ ∇ φ + ( u ∆ ∇ φ ) † � E ∇ = + 2 µ ∇ E ∇ 2 − 1 3 ( u ∆ · ∇ φ ) I u ∆ : slip velocity φ : volume fraction November 25, 2003 APS / DFD at New Jersey Page 5

Viscous Stress S – Results 5 0 4.5 E T 4 F 3.5 3 µ e 2.5 2 1.5 1 0.5 0 0 0.1 0.2 0.3 0.4 0.5 volume fraction φ 5 10 Part II Part II average average 8 4 ET ET FT FT FE FE 6 3 EM F2 4 µ∇ µ∆ 2 2 1 0 0 -2 -1 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 volume fraction φ volume fraction φ November 25, 2003 APS / DFD at New Jersey Page 6

Sedimentation Uniform suspensions — Sedimentation F u ∆ = U ( φ ) 6 πµ a u ∆ : slip velocity U ( φ ) : hindrance function F : interphase force Non-uniform suspensions — Shear F = 0 but u ∆ � 0 November 25, 2003 APS / DFD at New Jersey Page 7

Interphase Force F Closure relation F = F 1 u ∆ 6 πµ a F 2 a 2 E m · ∇ φ + F 3 a 2 ∇ 2 u m + F 4 a 2 ∇ × Ω ∆ + F 5 a 2 ( ∇ φ ) × Ω ∆ + F ⊥ a 2 � � ∇ 2 I − ∇∇ + · u ∆ d a 2 u ∆ · � � F ⊥ ∇ 2 I − ∇∇ + φ F � a 2 ∇∇ · u ∆ + d a 2 u ∆ · ( ∇∇ φ ) F � + November 25, 2003 APS / DFD at New Jersey Page 8

Interphase Force F – Results 35 0 30 -2 25 -4 20 -6 F 1 F 2 15 -8 10 -10 Part II 5 E|| -12 F ⊥ 0 0 0.1 0.2 0.3 0.4 0.5 0 0.1 0.2 0.3 0.4 0.5 volume fraction φ volume fraction φ 3 20 10 2 0 -10 1 -20 -30 0 -40 -50 -1 dF ⊥ /d φ F 1 /10 -60 F ⊥ F ⊥ d -2 -70 0 0.1 0.2 0.3 0.4 0.5 0 0.1 0.2 0.3 0.4 0.5 volume fraction φ volume fraction φ November 25, 2003 APS / DFD at New Jersey Page 9

Interphase Force F Closure relation Results and suggestions F 6 πµ a = F 1 u ∆ F 1 = 1 / U ( φ ) + F 2 a 2 E m · ∇ φ F 2 ≈ 2 d F 3 / d φ + F 3 a 2 ∇ 2 u m + F 4 a 2 ∇ × Ω ∆ + F 5 a 2 ( ∇ φ ) × Ω ∆ F 5 ≈ d F 4 / d φ + F ⊥ a 2 � F ⊥ ≈ F 1 / 10 for small φ � ∇ 2 I − ∇∇ · u ∆ d a 2 u ∆ · � � ∇ 2 I − ∇∇ + F ⊥ F ⊥ d ≈ d F ⊥ / d φ φ + F � a 2 ∇∇ · u ∆ F � = F 1 / 10 d a 2 u ∆ · ( ∇∇ φ ) + F � F � d = 0 November 25, 2003 APS / DFD at New Jersey Page 10

Discussions Expected constitutive equation of F : 1 + a 2 ∇ 2 � � F ( F 1 u ∆ ) 6 πµ a = 10 + a 2 ∇ · (2 F 3 E m ) + a 2 ∇ × ( F 4 Ω ∆ ) This suggests a relation between µ e and F 3 : 3 -2 F 3 µ e 2.5 2 1.5 1 0.5 0 0 0.1 0.2 0.3 0.4 0.5 . volume fraction φ November 25, 2003 APS / DFD at New Jersey Page 11

Conclusions • develop a systematic closure procedure for non-uniform suspensions • apply it to S and F • derive the constitutive equations, determine all closure coe ffi cients systematically, valid for both uniform and non-uniform suspensions and for all sedimentation, torque, and shear problems Future plans: • apply the closure procedure to interphase torque T and anti-symmetric part of the stress V • study the relation among the closure coe ffi cients November 25, 2003 APS / DFD at New Jersey Page 12

More Results of S S 2 µ e E m + 2 µ ∆ E ∆ + 2 µ ∇ E ∇ = µ � � + 2 µ 0 a 2 ∇ 2 E ∇ + 2 µ 1 a 2 E ∇ ∇ 2 φ 0.3 2.5 F in II Part II T in II 0.25 S in II 2 average 0.2 ET FT 0.15 1.5 FE µ 0 µ 1 0.1 1 0.05 0 0.5 -0.05 0 -0.1 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0 0.1 0.2 0.3 0.4 0.5 volume fraction φ volume fraction φ November 25, 2003 APS / DFD at New Jersey Page 13

More Results of F 0 4 -0.2 2 -0.4 0 -0.6 -2 F 3 -0.8 -4 -1 -6 -1.2 dF 3 /d φ -8 F 2 /2 E|| F 2 /2 F ⊥ -1.4 Part II -10 0 0.1 0.2 0.3 0.4 0.5 0 0.1 0.2 0.3 0.4 0.5 volume fraction φ volume fraction φ 1 4 0.5 2 0 -0.5 0 -1 -1.5 -2 -2 -2.5 -4 φ dF 4 /d φ -3 F 4 F 5 F 5 -3.5 -6 0 0.1 0.2 0.3 0.4 0.5 0 0.1 0.2 0.3 0.4 0.5 volume fraction φ volume fraction φ November 25, 2003 APS / DFD at New Jersey Page 14

Structure Factor S ( k ) 1.2 1 0.8 S(k) 0.6 0.4 0.2 0 1 2 3 4 5 6 k November 25, 2003 APS / DFD at New Jersey Page 15

Averages and Fitting 1 0.01 0.02 0.03 0.8 0.04 0.05 0.10 0.6 0.15 0.20 F [U] 0 0.25 0.30 0.4 0.35 0.40 0.45 0.2 0.50 0 0 0.2 0.4 0.6 0.8 1 1.2 k November 25, 2003 APS / DFD at New Jersey Page 16

Closure equations of F [ F ] 0 F 1 [ u ∆ ] 0 = F F 1 − k 2 � � � d F 1 � � F 1 − k 2 F � � [ F ] � [ u ∆ ] � d φ − k 2 F � [ u ∆ ] 0 = F + φ F F d 10 1 − k 2 � � � d F 1 � � F 1 − k 2 F ⊥ � d φ − k 2 F ⊥ [ u ∆ ] 0 [ F ] ⊥ [ u ∆ ] ⊥ = F + φ F d F 10 − F 3 k 2 [ u m ] ⊥ F + F 4 k [ Ω ∆ ] ⊥ F � F 1 − k 2 F ⊥ � [ F ] ⊥ [ u ∆ ] ⊥ T − F 3 k 2 [ u m ] ⊥ T + F 4 k [ Ω ∆ ] ⊥ = T T 1 − k 2 � � k [ Ω ∆ ] 0 + F 5 φ T 10 1 − k 2 � � � F 1 − k 2 F � � [ F ] � [ u ∆ ] � = E + F 2 k φ E 10 1 − k 2 � � � F 1 − k 2 F ⊥ � E − F 3 k 2 [ u m ] ⊥ [ F ] ⊥ [ u ∆ ] ⊥ − F 4 k [ Ω ∆ ] ⊥ = E + F 2 k φ E E 10 November 25, 2003 APS / DFD at New Jersey Page 17