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A Non-Newtonian rheology model application to complex flows (PyroClast)

Khawla Msheik, Meissa M’Baye, Duc Nguyen Supervisor: Francois James CEMRACS 2019 Luminy, August 22, 2019.

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Outline

Motivation Newtonian fluid (Known model)

Lubrication approximation Lubrication model Numerical results

Bingham fluid (Known model)

Navier-Stokes model for Bingham fluid Lubrication model Numerical results

Two viscosities model (New model)

Two-viscosities model for pseudo-viscoplastic fluid Lubrication model Numerical results

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Motivation

Pyroclast flow (Source: Internet)

Shear rate |Dv| Shear stress |τ| Newtonian pseudoplastic Bingham

Rheology - types Idea: To approximate the pseudo-plastic rheology by a piecewise affine function.

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Main features

H/L ≪ 1 Lubrication approximation Rheology - type (Stress-Strain relation).

Newtonian fluid : Water, oil,.... Bingham fluid Two - viscosities model: "Pseudo-viscoplastic fluid"

Navier-Stokes model

x z h

H L

u w

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Newtonian fluid

Navier-Stokes model

ux + wz = 0 ρ(ut + uux + wuz) = −px + 2µuxx + µuzz + µwxz ρ(wt + uwx + wwz) = −pz + µuxz + µwzz + 2µwzz − ρg Boundary condition u = w = 0 z = 0 (1 − h2

x)p + 2µ(1 + hx)2ux = 0

z = h(x, t) (1 − h2

x)(uz + wx) − 4hxux = 0

z = h(x, t) Kinematic condition ht + uhx = w z = h(x, t)

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Lubrication Method

Scale independent

x = L˜ x, z = H˜ z, u = U˜ u

Scale dependent

h = H˜ h, t = L U ˜ t, w = H LU ˜ w, p = ρgH˜ ρ ε = H L , Fr2 = U2 gH , Re = ρHU µ

Dimentionless equation

εRe(ut + uux + wuz) = −εRe Fr2 px + 2ε2µuxx + uzz + ε2wxz ε3Re(wt + uwx + wwz) = −εRe Fr2 pz + ε4µwxx + ε2uxz + 2ε2wzz − 1

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Lubrication model

Limit problem

ux + wz = 0, px = uzz, pz = −1 Boundary condition u = w = 0 z = 0 p = uz = 0 z = h(x, t) Kinematic condition ht + uhx = w z = h(x, t)

Lubrication model

ht − ∂x h3 3 hx

• = 0

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Numerical results

Method 1: F(Hl, Hr) = − HL + HR 2 3 HR − HL 3∆x Method 2: F(Hl, Hr) = (HL − HR)(H3

L + H3 R)

6∆x − |HL − HR| max(H2

L, H2 R)(HR − HL)

2∆x

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Bingham fluid

v = (u, w)T Dv = ∇v+∇vT

2

, τ = τxx τxz τxz τzz

• Laws of behavior

Newtonian fluid τ = 2µDv Bingham fluid        |τ| < Bi Dv = 0 solid τ = Dv + Bi Dv

|Dv|

Dv 0 fluid

Limit problem

Newtonian fluid px = uzz Bingham fluid px = 2∂zτxz = ∂z      uz      1 + Bi √ 2 |uz|            

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Lubrication model - Numerical results

ht − ∂x hxY 2 6 [3h − Y]

• = 0

Y = max      h − Bi √ 2 |hx| , 0       Remark: Bi = 0 Bingham → Newtonian Bi = 0.5 Bi = 2 nx = 128

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Numerical results - Slump test

Bi = 2 Bi = 1.25 Bi = 0.5

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Two-viscosities model τ =        2µ1Dv |Dv| ≤

τ∗ 2µ1

high viscosity 2µ2Dv +

• 1 − µ2

µ1

• τ∗ Dv

|Dv|

|Dv| >

τ∗ 2µ1

low viscosity µ2 = µ1 τ = 2µ1Dv Newtonian fluid µ1 → +∞ τ =        |τ| < τ∗ |Dv| ≤ 0 2µ2Dv + τ∗ Dv

|Dv|

|Dv| > 0 Bingham fluid |Dv| |τ| µ2 → µ1 τ∗ µ1 → ∞ µ

1

µ

2 τ∗ 2µ1

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Two-viscosities model

low viscosity, u2 high viscosity, u1 h∗ h

Difficulty

1

Induce unknown interface h∗

2

Identify BC at interface h∗

Resolving

1

Choice of scaling + constraint on |Du|

2

Preserve continuity of u and τ

3

Scale the viscosity: µi = αiµ

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Lubrication model

Two-viscosities model ∂th − ∂x

• hx

1 µ2 h∗3 3 − Yh∗2 2 + Yhh∗ − hh∗2 2

• − 1

µ1 h∗3 − h3 3 + h2h∗ − hh∗2 = 0 Y = max      h −

• 1 − µ2

µ1 √ 2B |hx| , 0       , h∗ = max      h − √ 2B |hx| , 0       µ2 = µ1 ∂th − ∂x h3 3 hx

• = 0

Newtonian fluid µ1 → +∞

ht − ∂x hxY 2 6 [3h − Y]

• = 0

Y = max      h − Bi √ 2 |hx| , 0       Bingham fluid

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Numerical results

µ2 = µ1 Newtonian fluid µ1 = 1000 ≫ µ2 Bingham fluid

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Slump test for two-viscosities fluid

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Thanks CEMRACS, 2019

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