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

SLIDE 1

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)

SLIDE 2

Introduction

Non-uniform suspensions

Practically important:

• Shear-induced diffusivity
• Particle migration in Stokes flows
• Stratification in sedimentation

Uniform suspension is too simple:

• No strain in sedimentation
• No slip velocity in shear problem

Important physics vanishes!

.

k ^

φ φ

November 25, 2003 APS/DFD at New Jersey Page 2

SLIDE 3

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

SLIDE 4

Rheology Uniform suspensions — Shear

S

µ = 2 µe Em

Em

= 1 2

• ∇um + (∇um)†

um : mixture velocity µ : viscosity of the fluid µe : relative viscosity of the mixture

Non-uniform suspensions – Sedimentation

Em 0 and

µe plays a role

November 25, 2003 APS/DFD at New Jersey Page 4

SLIDE 5

Viscous Stress S Closure relation

S

µ = 2 µe Em + 2 µ∆ E∆ + 2 µ∇ E∇

E∆

= 1 2

• ∇u∆ + (∇u∆)†

−1 3 (∇ · u∆) I

E∇

= 1 2

• u∆∇φ + (u∆∇φ)†

−1 3 (u∆ · ∇φ) I u∆ : slip velocity φ : volume fraction

November 25, 2003 APS/DFD at New Jersey Page 5

SLIDE 6

Viscous Stress S – Results

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0.1 0.2 0.3 0.4 0.5

µe

volume fraction φ E T F

• 1

1 2 3 4 5 0.05 0.1 0.15 0.2 0.25 0.3 0.35

µ∆

volume fraction φ Part II average ET FT FE EM F2

• 2

2 4 6 8 10 0.05 0.1 0.15 0.2 0.25 0.3 0.35

µ∇

volume fraction φ Part II average ET FT FE

November 25, 2003 APS/DFD at New Jersey Page 6

SLIDE 7

Sedimentation Uniform suspensions — Sedimentation u∆ = U(φ) F 6πµa

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

SLIDE 8

Interphase Force F

Closure relation

F 6πµa = F1 u∆ + F2 a2 Em · ∇φ + F3 a2∇2um + F4 a2 ∇ × Ω∆ + F5 a2 (∇φ) × Ω∆ + F⊥ a2 ∇2I − ∇∇

• · u∆

+ F⊥

d a2 u∆ ·

• ∇2I − ∇∇
• φ

+ F a2 ∇∇ · u∆ + F

d a2 u∆ · (∇∇φ)

November 25, 2003 APS/DFD at New Jersey Page 8

SLIDE 9

Interphase Force F – Results

5 10 15 20 25 30 35 0.1 0.2 0.3 0.4 0.5 F1 volume fraction φ

• 12
• 10
• 8
• 6
• 4
• 2

0.1 0.2 0.3 0.4 0.5 F2 volume fraction φ Part II E|| F⊥

• 2
• 1

1 2 3 0.1 0.2 0.3 0.4 0.5 volume fraction φ F1/10 F⊥

• 70
• 60
• 50
• 40
• 30
• 20
• 10

10 20 0.1 0.2 0.3 0.4 0.5 volume fraction φ dF⊥/dφ F⊥

d

November 25, 2003 APS/DFD at New Jersey Page 9

SLIDE 10

Interphase Force F

Closure relation

Results and suggestions F 6πµa = F1 u∆ F1 = 1/U(φ) + F2 a2 Em · ∇φ F2 ≈ 2 dF3/dφ + F3 a2∇2um + F4 a2 ∇ × Ω∆ + F5 a2 (∇φ) × Ω∆ F5 ≈ dF4/dφ + F⊥ a2 ∇2I − ∇∇

• · u∆

F⊥ ≈ F1/10 for small φ + F⊥

d a2 u∆ ·

• ∇2I − ∇∇
• φ

F⊥

d ≈ dF⊥/dφ

+ F a2 ∇∇ · u∆ F = F1/10 + F

d a2 u∆ · (∇∇φ)

F

d = 0

November 25, 2003 APS/DFD at New Jersey Page 10

SLIDE 11

Discussions

Expected constitutive equation of F : F 6πµa =

• 1 + a2∇2

10

• (F1 u∆)

+ a2∇ · (2 F3 Em) + a2∇ × (F4 Ω∆)

This suggests a relation between µe and F3: .

0.5 1 1.5 2 2.5 3 0.1 0.2 0.3 0.4 0.5 volume fraction φ

• 2 F3

µe

November 25, 2003 APS/DFD at New Jersey Page 11

SLIDE 12

Conclusions

• develop a systematic closure procedure

for non-uniform suspensions

• apply it to S and F
• derive the constitutive equations,

determine all closure coefficients 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 coefficients

November 25, 2003 APS/DFD at New Jersey Page 12

SLIDE 13

More Results of S

S

µ = 2 µe Em + 2 µ∆ E∆ + 2 µ∇ E∇ +2 µ0 a2∇2E∇ + 2 µ1 a2E∇

• ∇2φ
• 0.1
• 0.05

0.05 0.1 0.15 0.2 0.25 0.3 0.05 0.1 0.15 0.2 0.25 0.3 0.35

µ0

volume fraction φ F in II T in II S in II average ET FT FE 0.5 1 1.5 2 2.5 0.1 0.2 0.3 0.4 0.5

µ1

volume fraction φ Part II

November 25, 2003 APS/DFD at New Jersey Page 13

SLIDE 14

More Results of F

• 10
• 8
• 6
• 4
• 2

2 4 0.1 0.2 0.3 0.4 0.5 volume fraction φ dF3/dφ F2/2 E|| F2/2 F⊥

• 1.4
• 1.2
• 1
• 0.8
• 0.6
• 0.4
• 0.2

0.1 0.2 0.3 0.4 0.5 F3 volume fraction φ Part II

• 3.5
• 3
• 2.5
• 2
• 1.5
• 1
• 0.5

0.5 1 0.1 0.2 0.3 0.4 0.5 volume fraction φ F4 F5

• 6
• 4
• 2

2 4 0.1 0.2 0.3 0.4 0.5 volume fraction φ φ dF4/dφ F5

November 25, 2003 APS/DFD at New Jersey Page 14

SLIDE 15

Structure Factor S (k)

0.2 0.4 0.6 0.8 1 1.2 1 2 3 4 5 6

S(k) k

November 25, 2003 APS/DFD at New Jersey Page 15

SLIDE 16

Averages and Fitting

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 1.2 [U]0

F

k

0.01 0.02 0.03 0.04 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

November 25, 2003 APS/DFD at New Jersey Page 16

SLIDE 17

Closure equations of F

[F]0

F

= F1[u∆]0

F

[F]

F

=

• F1 − k2F

[u∆]

F + φ

• 1 − k2

10 dF1 dφ − k2F

d

• [u∆]0

F

[F]⊥

F

=

• F1 − k2F⊥

[u∆]⊥

F + φ

• 1 − k2

10 dF1 dφ − k2F⊥

d

• [u∆]0

F

−F3k2[um]⊥

F + F4k[Ω∆]⊥ F

[F]⊥

T

=

• F1 − k2F⊥

[u∆]⊥

T − F3k2[um]⊥ T + F4k[Ω∆]⊥ T

+F5φ

• 1 − k2

10

• k[Ω∆]0

T

[F]

E

=

• F1 − k2F

[u∆]

E + F2kφ

• 1 − k2

10

• [F]⊥

E

=

• F1 − k2F⊥

[u∆]⊥

E − F3k2[um]⊥ E + F2kφ

• 1 − k2

10

• − F4k[Ω∆]⊥

E

November 25, 2003 APS/DFD at New Jersey Page 17