A Non-Newtonian rheology model application to complex flows - PowerPoint PPT Presentation

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. Khawla, Meissa, Duc CEMRACS 2019 1 / 17

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 Khawla, Meissa, Duc CEMRACS 2019 2 / 17

Motivation Bingham Shear stress | τ | Newtonian pseudoplastic Shear rate | Dv | Pyroclast flow (Source: Internet) Rheology - types Idea: To approximate the pseudo-plastic rheology by a piecewise affine function. Khawla, Meissa, Duc CEMRACS 2019 3 / 17

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 L w u H z h x Khawla, Meissa, Duc CEMRACS 2019 4 / 17

Newtonian fluid Navier-Stokes model u x + w z = 0 ρ ( u t + uu x + wu z ) = − p x + 2 µ u xx + µ u zz + µ w xz ρ ( w t + uw x + ww z ) = − p z + µ u xz + µ w zz + 2 µ w zz − ρ g Boundary condition u = w = 0 z = 0 ( 1 − h 2 x ) p + 2 µ ( 1 + h x ) 2 u x = 0 z = h ( x , t ) ( 1 − h 2 x )( u z + w x ) − 4 h x u x = 0 z = h ( x , t ) Kinematic condition h t + uh x = w z = h ( x , t ) Khawla, Meissa, Duc CEMRACS 2019 5 / 17

Lubrication Method Scale independent x = L ˜ z = H ˜ u = U ˜ x , z , u Scale dependent t = L w = H h = H ˜ ˜ LU ˜ p = ρ gH ˜ h , t , w , ρ U Fr 2 = U 2 ε = H Re = ρ HU L , gH , µ Dimentionless equation ε Re ( u t + uu x + wu z ) = − ε Re Fr 2 p x + 2 ε 2 µ u xx + u zz + ε 2 w xz ε 3 Re ( w t + uw x + ww z ) = − ε Re Fr 2 p z + ε 4 µ w xx + ε 2 u xz + 2 ε 2 w zz − 1 Khawla, Meissa, Duc CEMRACS 2019 6 / 17

Lubrication model Limit problem u x + w z = 0 , p x = u zz , p z = − 1 Boundary condition u = w = 0 z = 0 p = u z = 0 z = h ( x , t ) Kinematic condition h t + uh x = w z = h ( x , t ) Lubrication model � h 3 � = 0 h t − ∂ x 3 h x Khawla, Meissa, Duc CEMRACS 2019 7 / 17

Numerical results Method 1: � 3 H R − H L � H L + H R F ( H l , H r ) = − 3 ∆ x 2 Method 2: F ( H l , H r ) = ( H L − H R )( H 3 L + H 3 − | H L − H R | max( H 2 L , H 2 R ) R )( H R − H L ) 6 ∆ x 2 ∆ x Khawla, Meissa, Duc CEMRACS 2019 8 / 17

Bingham fluid v = ( u , w ) T � τ xx � τ xz Dv = ∇ v + ∇ v T τ = , 2 τ xz τ zz Laws of behavior Newtonian fluid Bingham fluid  τ = 2 µ Dv | τ | < Bi Dv = 0 solid    τ = Dv + Bi Dv  Dv � 0 fluid   | Dv | Limit problem Newtonian fluid Bingham fluid √ p x = u zz      1 + Bi 2 p x = 2 ∂ z τ xz = ∂ z      u z             | u z |   Khawla, Meissa, Duc CEMRACS 2019 9 / 17

Lubrication model - Numerical results � h x Y 2 � h t − ∂ x [ 3 h − Y ] = 0 6 √    h − Bi 2   Y = max | h x | , 0        Remark: Bi = 0 Bingham → Newtonian Bi = 0 . 5 Bi = 2 nx = 128 Khawla, Meissa, Duc CEMRACS 2019 10 / 17

Numerical results - Slump test Bi = 2 Bi = 1 . 25 Bi = 0 . 5 Khawla, Meissa, Duc CEMRACS 2019 11 / 17

Two-viscosities model  τ ∗ 2 µ 1 Dv | Dv | ≤ high viscosity  2 µ 1  τ =  � 1 − µ 2 � τ ∗ τ ∗ Dv 2 µ 2 Dv +  | Dv | > low viscosity   µ 1 | Dv | 2 µ 1 µ 2 = µ 1 µ 1 → + ∞  τ = 2 µ 1 Dv | τ | < τ ∗ | Dv | ≤ 0   τ =  2 µ 2 Dv + τ ∗ Dv  | Dv | > 0   | Dv | Newtonian fluid Bingham fluid | τ | µ 2 → µ 1 µ 1 → ∞ µ 2 τ ∗ µ 1 τ ∗ | Dv | 2 µ 1 Khawla, Meissa, Duc CEMRACS 2019 12 / 17

Two-viscosities model high viscosity, u 1 h h ∗ low viscosity, u 2 Difficulty Resolving Induce unknown interface h ∗ Choice of scaling + constraint on | Du | 1 1 Identify BC at interface h ∗ Preserve continuity of u and τ 2 2 Scale the viscosity: µ i = α i µ 3 Khawla, Meissa, Duc CEMRACS 2019 13 / 17

Lubrication model Two-viscosities model � 1 � h ∗ 3 − h 3 � � h ∗ 3 3 − Yh ∗ 2 + Yhh ∗ − hh ∗ 2 + h 2 h ∗ − hh ∗ 2 �� � − 1 � ∂ t h − ∂ x h x = 0 µ 2 2 2 µ 1 3 � √ √     � 2 B 2 B 1 − µ 2 h ∗ = max     Y = max  h − | h x | , 0  ,  h − | h x | , 0             µ 1  µ 2 = µ 1 µ 1 → + ∞ � h x Y 2 � [ 3 h − Y ] = 0 h t − ∂ x � h 3 � 6 ∂ t h − ∂ x 3 h x = 0 √    h − Bi 2   Y = max | h x | , 0       Newtonian fluid  Bingham fluid Khawla, Meissa, Duc CEMRACS 2019 14 / 17

Numerical results µ 2 = µ 1 µ 1 = 1000 ≫ µ 2 Newtonian fluid Bingham fluid Khawla, Meissa, Duc CEMRACS 2019 15 / 17

Slump test for two-viscosities fluid Khawla, Meissa, Duc CEMRACS 2019 16 / 17

Thanks CEMRACS, 2019 Khawla, Meissa, Duc CEMRACS 2019 17 / 17