Non-homogeneous incompressible Bingham flows withvariable yield stress and application to volcanology.

Jordane Mathéwith Laurent Chupin (maths) and

Karim Kelfoun (LMV)

Laboratoire de Maths & Laboratoire Magmas et Volcans

june 2015

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 1 / 18

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 2 / 18

Laboratory experiment

Zoom into the granular column →

O. Roche, LMV.

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 3 / 18

Laboratory experiment

Zoom into the granular column →

Fluidisation :injection of gas

througha pore plate.

O. Roche, LMV.

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 3 / 18

Figure: Non-fluidised → short runout distance

Figure: Fluidised → long runout distance

1 Model

2 Numerical simulation

3 Perspectives

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 5 / 18

1 Model

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 6 / 18

Two phases for one fluid

Consider one mixed fluid with variable density.

It satisfies thenon-homogeneous incompressible Navier-Stokes equations:

Two phases for one fluid

Consider one mixed fluid with variable density. It satisfies thenon-homogeneous incompressible Navier-Stokes equations:

div(v) = 0∂tρ+ div(ρv) = 0∂t(ρv) + div(ρv ⊗ v) +∇p = ρg + div(S)

Two phases for one fluid

Consider one mixed fluid with variable density. It satisfies thenon-homogeneous incompressible Navier-Stokes equations:

div(v) = 0∂tρ+ div(ρv) = 0∂t(ρv) + div(ρv ⊗ v) +∇p = ρg + div(S)

Unknown:v : velocity,p: total pressure,ρ: density.

Two phases for one fluid

Consider one mixed fluid with variable density. It satisfies thenon-homogeneous incompressible Navier-Stokes equations:

div(v) = 0∂tρ+ div(ρv) = 0∂t(ρv) + div(ρv ⊗ v) +∇p = ρg + div(S)

Constantg : gravity.

Two phases for one fluid

Consider one mixed fluid with variable density. It satisfies thenon-homogeneous incompressible Navier-Stokes equations:

div(v) = 0∂tρ+ div(ρv) = 0∂t(ρv) + div(ρv ⊗ v) +∇p = ρg + div(S)

Constantg : gravity.

RheologyLet precise S.

RheologyIn our case

S = µDv +

Σ ,

where Dv = γ is the strain rate tensor,µ is the effective viscosity

Σ =q Dv|Dv |

⇒ Bingham fluid with yield stress q

IdeaLet vary the yield stress q as a function of the interstitial gas pressure.

RheologyIn our case

S = µDv + Σ ,

where Dv = γ is the strain rate tensor,µ is the effective viscosity Σ =q Dv

|Dv |

⇒ Bingham fluid with yield stress q

IdeaLet vary the yield stress q as a function of the interstitial gas pressure.

RheologyIn our case

S = µDv + Σ ,

where Dv = γ is the strain rate tensor,µ is the effective viscosity Σ =q Dv

|Dv |

⇒ Bingham fluid with yield stress q

IdeaLet vary the yield stress q as a function of the interstitial gas pressure.

Variation of the yield stress

Definition of the yield stress

q =

atmospheric pressure: Coulomb frictionhigh pressure: fluid

q =

tan(δ)(ρgh − gas pressure) if low gas pressure0 if high gas pressure

q = tan(δ)(ρgh − gas pressure)+

where δ is the internal friction angle.

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 9 / 18

Variation of the yield stress

Definition of the yield stress

q =

atmospheric pressure: Coulomb frictionhigh pressure: fluid

q =

tan(δ)(ρgh − gas pressure) if low gas pressure0 if high gas pressure

q = tan(δ)(ρgh − gas pressure)+

where δ is the internal friction angle.

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 9 / 18

Variation of the yield stress

Definition of the yield stress

q =

atmospheric pressure: Coulomb frictionhigh pressure: fluid

q =

tan(δ)(ρgh − gas pressure) if low gas pressure0 if high gas pressure

q = tan(δ)(ρgh − gas pressure)+

where δ is the internal friction angle.

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 9 / 18

Equations of the model

Noting pf the interstitial gas pressure, we obtain:

div(v) = 0∂tρ+ div(ρv) = 0

∂t(ρv) + div(ρv ⊗ v)− µ∆v +∇p = tan(δ) div

(ρgh−pf )+︸ ︷︷ ︸yield q

Dv|Dv |

+ ρg

∂tpf + v · ∇pf − κ∆pf = 0

where κ is the diffusion coefficient.

Equations of the model

Noting pf the interstitial gas pressure, we obtain:

div(v) = 0∂tρ+ div(ρv) = 0

∂t(ρv) + div(ρv ⊗ v)− µ∆v +∇p = tan(δ) div

(ρgh−pf )+︸ ︷︷ ︸yield q

Dv|Dv |

+ ρg

∂tpf + v · ∇pf − κ∆pf = 0

where κ is the diffusion coefficient.

2 Numerical simulation

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 11 / 18

Dambreak: Mesh

∂tρ+ v · ∇ρ = 0

numerical method:RK3 TVD - WENO5 scheme.

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 12 / 18

Dambreak: how to treat the density?

∂tρ+ v · ∇ρ = 0

numerical method:RK3 TVD - WENO5 scheme.

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 12 / 18

Numerical scheme (without density)n = 0 v0, Σ0, p0 and pf

0 given.

n > 0 vn, Σn, pn and pfn being calculated.

We compute (vn+1,Σn+1) solution of3vn+1 − 4vn + vn−1

2δt + 2vn ·∇vn − vn−1 ·∇vn−1 − ∆vn+1

ρn+1 + ∇pn

ρn+1 = div Σn+1

ρn+1 − ey ,

Σn+1 = Pqn (Σn+1 + r Dvn+1 + ε(Σn − Σn+1)),

vn+1∣∣∣∂Ω

= 0.

→ k = 0 Σn,0 = Σn, and (vn, pn, pnf ) given.

→ k > 0 Σn,k known, we first compute vn,k solution of aLaplace-type problem then we project the stress tensor to obtain Σn,k+1:

3vn,k − 4vn + vn−1

2δt + 2vn ·∇vn − vn−1 ·∇vn−1 − ∆vn,k

ρn+1 + ∇pn

ρn+1 = div Σn,k

ρn+1 − ey ,

vn,k∣∣∣∂Ω

= 0,

Σn,k+1 = Pqn (Σn,k + r Dvn,k + ε(Σn − Σn,k )).

Numerical scheme (without density)n = 0 v0, Σ0, p0 and pf

0 given.n > 0 vn, Σn, pn and pf

n being calculated.

We compute (vn+1,Σn+1) solution of3vn+1 − 4vn + vn−1

2δt + 2vn ·∇vn − vn−1 ·∇vn−1 − ∆vn+1

ρn+1 + ∇pn

ρn+1 = div Σn+1

ρn+1 − ey ,

Σn+1 = Pqn (Σn+1 + r Dvn+1 + ε(Σn − Σn+1)),

vn+1∣∣∣∂Ω

= 0.

→ k = 0 Σn,0 = Σn, and (vn, pn, pnf ) given.

→ k > 0 Σn,k known, we first compute vn,k solution of aLaplace-type problem then we project the stress tensor to obtain Σn,k+1:

3vn,k − 4vn + vn−1

2δt + 2vn ·∇vn − vn−1 ·∇vn−1 − ∆vn,k

ρn+1 + ∇pn

ρn+1 = div Σn,k

ρn+1 − ey ,

vn,k∣∣∣∂Ω

= 0,

Σn,k+1 = Pqn (Σn,k + r Dvn,k + ε(Σn − Σn,k )).

Numerical scheme (without density)n = 0 v0, Σ0, p0 and pf

0 given.n > 0 vn, Σn, pn and pf

n being calculated.We compute (vn+1,Σn+1) solution of

3vn+1 − 4vn + vn−1

2δt + 2vn ·∇vn − vn−1 ·∇vn−1 − ∆vn+1

ρn+1 + ∇pn

ρn+1 = div Σn+1

ρn+1 − ey ,

Σn+1 = Pqn (Σn+1 + r Dvn+1 + ε(Σn − Σn+1)),

vn+1∣∣∣∂Ω

= 0.

→ k = 0 Σn,0 = Σn, and (vn, pn, pnf ) given.

→ k > 0 Σn,k known, we first compute vn,k solution of aLaplace-type problem then we project the stress tensor to obtain Σn,k+1:

3vn,k − 4vn + vn−1

2δt + 2vn ·∇vn − vn−1 ·∇vn−1 − ∆vn,k

ρn+1 + ∇pn

ρn+1 = div Σn,k

ρn+1 − ey ,

vn,k∣∣∣∂Ω

= 0,

Σn,k+1 = Pqn (Σn,k + r Dvn,k + ε(Σn − Σn,k )).

Numerical scheme (without density)n = 0 v0, Σ0, p0 and pf

0 given.n > 0 vn, Σn, pn and pf

n being calculated.We compute (vn+1,Σn+1) solution of

3vn+1 − 4vn + vn−1

2δt + 2vn ·∇vn − vn−1 ·∇vn−1 − ∆vn+1

ρn+1 + ∇pn

ρn+1 = div Σn+1

ρn+1 − ey ,

Σn+1 = Pqn (Σn+1 + r Dvn+1 + ε(Σn − Σn+1)),

vn+1∣∣∣∂Ω

= 0.

We compute (vn+1, pn+1) thanks to the incompressibility constrainvn+1 − vn+1

δt + 23ρn+1∇(pn+1 − pn) = 0,

div vn+1 = 0, vn+1 · n∣∣∣∂Ω

= 0.

→ k = 0 Σn,0 = Σn, and (vn, pn, pnf ) given.

→ k > 0 Σn,k known, we first compute vn,k solution of aLaplace-type problem then we project the stress tensor to obtain Σn,k+1:

3vn,k − 4vn + vn−1

2δt + 2vn ·∇vn − vn−1 ·∇vn−1 − ∆vn,k

ρn+1 + ∇pn

ρn+1 = div Σn,k

ρn+1 − ey ,

vn,k∣∣∣∂Ω

= 0,

Σn,k+1 = Pqn (Σn,k + r Dvn,k + ε(Σn − Σn,k )).

Numerical scheme (without density)n = 0 v0, Σ0, p0 and pf

0 given.n > 0 vn, Σn, pn and pf

n being calculated.We compute (vn+1,Σn+1) solution of

3vn+1 − 4vn + vn−1

2δt + 2vn ·∇vn − vn−1 ·∇vn−1 − ∆vn+1

ρn+1 + ∇pn

ρn+1 = div Σn+1

ρn+1 − ey ,

Σn+1 = Pqn (Σn+1 + r Dvn+1 + ε(Σn − Σn+1)),

vn+1∣∣∣∂Ω

= 0.

We compute (vn+1, pn+1) thanks to the incompressibility constrainvn+1 − vn+1

δt + 23ρn+1∇(pn+1 − pn) = 0,

div vn+1 = 0, vn+1 · n∣∣∣∂Ω

= 0.

We compute pfn+1 solution of

3pfn+1 − 4pf

n + pfn−1

2δt + 2vn+1 · ∇pfn − vn+1 · ∇pf

n−1 −∆pfn+1 = 0 .

→ k = 0 Σn,0 = Σn, and (vn, pn, pnf ) given.

→ k > 0 Σn,k known, we first compute vn,k solution of aLaplace-type problem then we project the stress tensor to obtain Σn,k+1:

3vn,k − 4vn + vn−1

2δt + 2vn ·∇vn − vn−1 ·∇vn−1 − ∆vn,k

ρn+1 + ∇pn

ρn+1 = div Σn,k

ρn+1 − ey ,

vn,k∣∣∣∂Ω

= 0,

Σn,k+1 = Pqn (Σn,k + r Dvn,k + ε(Σn − Σn,k )).

Numerical scheme (without density)n = 0 v0, Σ0, p0 and pf

0 given.n > 0 vn, Σn, pn and pf

n being calculated.We compute (vn+1,Σn+1) solution of

3vn+1 − 4vn + vn−1

2δt + 2vn ·∇vn − vn−1 ·∇vn−1 − ∆vn+1

ρn+1 + ∇pn

ρn+1 = div Σn+1

ρn+1 − ey ,

Σn+1 = Pqn (Σn+1 + r Dvn+1 + ε(Σn − Σn+1)),

vn+1∣∣∣∂Ω

= 0.

→ k = 0 Σn,0 = Σn, and (vn, pn, pnf ) given.

→ k > 0 Σn,k known, we first compute vn,k solution of aLaplace-type problem then we project the stress tensor to obtain Σn,k+1:

3vn,k − 4vn + vn−1

2δt + 2vn ·∇vn − vn−1 ·∇vn−1 − ∆vn,k

ρn+1 + ∇pn

ρn+1 = div Σn,k

ρn+1 − ey ,

vn,k∣∣∣∂Ω

= 0,

Σn,k+1 = Pqn (Σn,k + r Dvn,k + ε(Σn − Σn,k )).

Numerical scheme (without density)n = 0 v0, Σ0, p0 and pf

0 given.n > 0 vn, Σn, pn and pf

n being calculated.We compute (vn+1,Σn+1) solution of

3vn+1 − 4vn + vn−1

2δt + 2vn ·∇vn − vn−1 ·∇vn−1 − ∆vn+1

ρn+1 + ∇pn

ρn+1 = div Σn+1

ρn+1 − ey ,

Σn+1 = Pqn (Σn+1 + r Dvn+1 + ε(Σn − Σn+1)),

vn+1∣∣∣∂Ω

= 0.

→ k = 0 Σn,0 = Σn, and (vn, pn, pnf ) given.

→ k > 0 Σn,k known, we first compute vn,k solution of aLaplace-type problem then we project the stress tensor to obtain Σn,k+1:

3vn,k − 4vn + vn−1

2δt + 2vn ·∇vn − vn−1 ·∇vn−1 − ∆vn,k

ρn+1 + ∇pn

ρn+1 = div Σn,k

ρn+1 − ey ,

vn,k∣∣∣∂Ω

= 0,

Σn,k+1 = Pqn (Σn,k + r Dvn,k + ε(Σn − Σn,k )).

Numerical scheme (without density)n = 0 v0, Σ0, p0 and pf

0 given.n > 0 vn, Σn, pn and pf

n being calculated.We compute (vn+1,Σn+1) solution of

3vn+1 − 4vn + vn−1

2δt + 2vn ·∇vn − vn−1 ·∇vn−1 − ∆vn+1

ρn+1 + ∇pn

ρn+1 = div Σn+1

ρn+1 − ey ,

Σn+1 = Pqn (Σn+1 + r Dvn+1 + ε(Σn − Σn+1)),

vn+1∣∣∣∂Ω

= 0.

→ k = 0 Σn,0 = Σn, and (vn, pn, pnf ) given.

→ k > 0 Σn,k known, we first compute vn,k solution of aLaplace-type problem then we project the stress tensor to obtain Σn,k+1:

3vn,k − 4vn + vn−1

2δt + 2vn ·∇vn − vn−1 ·∇vn−1 − ∆vn,k

ρn+1 + ∇pn

ρn+1 = div Σn,k

ρn+1 − ey ,

vn,k∣∣∣∂Ω

= 0,

Σn,k+1 = Pqn (Σn,k + r Dvn,k + ε(Σn − Σn,k )).

Experimental conditions

14cm

20 cm

ρ = 1550 kg.m−3

µ = 1 Pa.sδ = 34 degrees

density: 1 kg.m−3

viscosity: 1 Pa.s

granular bed “air”

Figure: Experimental setup

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 14 / 18

Numerical simulation

We compute the collapse with different diffusion coefficient κ.

With κ = 1, it happens nothing.κ = 0.2κ = 0.1

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 15 / 18

Diffusion coefficient = 0.2

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 15 / 18

Diffusion coefficient = 0.1

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 15 / 18

3 Perspectives

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 16 / 18

Perspectives

To compare the results given by this numerical scheme for thenon-fluidised case:

I Lower yield stress,. . .

I Lower friction coefficient.. . .

To compare the results given by this numerical scheme for thefluidised case:

I Adapt the viscosity value.

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 17 / 18

Thank you!

Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 18 / 18