Non-homogeneous incompressible Bingham flows with · PDF fileLaboratoryexperiment...
Transcript of Non-homogeneous incompressible Bingham flows with · PDF fileLaboratoryexperiment...
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
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Laboratory experiment
Zoom into the granular column →
O. Roche, LMV.
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Laboratory experiment
Zoom into the granular column →
Fluidisation :injection of gas
througha pore plate.
O. Roche, LMV.
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Figure: Non-fluidised → short runout distance
Figure: Fluidised → long runout distance
1 Model
2 Numerical simulation
3 Perspectives
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1 Model
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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
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Dambreak: Mesh
∂tρ+ v · ∇ρ = 0
numerical method:RK3 TVD - WENO5 scheme.
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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
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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
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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.
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Thank you!
Jordane Mathé (Maths-LMV) Fluidized granular flows june 2015 18 / 18