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Two-phase hydrodynamic model for air entrainment at moving contact line Tak Shing Chan and Jacco Snoeijer Physics of Fluids Group Faculty of Science and Technology University of Twente Part one: Introduction Introduction: air Static contact angle θo liquid Introduction: Constant U Dewetting U Ca (receding contact air line): liquid Introduction: U > Uc e.g. Landau-Levich- Dewetting Derjaguin film U Ca (receding contact air line): Cac~10-2 liquid Bonn et al. (Rev. Mod. Phys. 2009) Lubrication theory Introduction: Wetting U Ca (advancing contact air line): liquid Constant U Introduction: Wetting U Ca (advancing contact air line): liquid Air entrainment ? U > Uc Instability of advancing contact line (experimental motivation) A fiber is pulled into a liquid bath. Pressurized liquid, Cac ~ 50 (P.G. Simpkins & V.J. Kuck, J. Colloid & Interface Sci. 263, 2003) A splash is observed when the Dip coating: air bubbles are speed of the bead is larger observed. Cac ~1 than a threshold value. (H. Benkreira & M.I. Khan, (Duez, C. et al Nature Phys. Chem. Engineering Sci. 63, 3, 2007) 2008) Questions: Introduction: What is the mechanism for air entrainment? Can we compute the critical Cac theoretically? Wetting U Ca (advancing contact air line): liquid U > Uc Questions: Introduction: What is the mechanism for air entrainment? Can we compute the critical Cac theoretically? Wetting U Ca (advancing contact air line): liquid Lubrication theory still valid ??? U > Uc Air flow important ??? Analogy with free surface cusp: role of air flow Lorenceau, Restagno, Quere, PRL 2003 Eggers PRL 2001 air Increasing speed liquid critical Ca depends on viscosity ratio !! Analogy with free surface cusp: role of air flow Lorenceau, Restagno, Quere, PRL 2003 Eggers PRL 2001 air Increasing speed liquid critical Ca depends on viscosity ratio !! What happens for flow with a contact line? Part two: 2-phase hydrodynamic model 2-phase model: Assume straight contact line (2D problem) h We consider very small Re number (Re << 1)and stationary state ( 0) only: t Fluid A (e.g. air) interface h Fluid B (e.g. water) Constant speed U 2-phase model: Assume straight contact line (2D problem) h We consider very small Re number (Re << 1)and stationary state ( 0) only: t Fluid A (e.g. air) interface h Young-Laplace equation PA PB Fluid B (e.g. water) Constant speed U 2-phase model: Assume straight contact line (2D problem) h We consider very small Re number (Re << 1)and stationary state ( 0) only: t Fluid A (e.g. air) interface h Young-Laplace equation PA PB Fluid B (e.g. water) Stokes equation (Re<< 1) P U gravity 2 Constant speed U 2-phase model: For standard lubrication theory (1 phase, small slope), we use Poiseuille flow to approximate the velocity field. h x d 3h 3Ca dh dx 3 h( h 3 ) dx 2-phase model: For standard lubrication theory (1 phase, small slope), we use Poiseuille flow to approximate the velocity field. h x d 3h 3Ca dh dx 3 h( h 3 ) dx For two phase flow ??? Huh & Scriven’s solution in straight wedge problem Stream lines air liquid θ U (C. Huh & L.E. Scriven, Journal of Colloid and Interface Science, 1971). 2-phase model: Our idea is… 1 1 2 2 3 …… 3 With the assumption that the curvature of interface is small, we approximate the flow in our wetting problem by the flow in straight wedge problem. 2-phase model: Fluid A (e.g. air) d 2 3Ca B f ( , R ) cos h interface ds 2 h(h 3 ) θ Fluid B (e.g. water) U 2 sin 3 [ R 2 ( 2 sin 2 ) 2 R{ ( ) sin 2 } {( ) 2 sin 2 }] f ( , R ) 3[ R (sin 2 2 ){( ) sin cos } {sin 2 ( ) 2 }( sin cos )] 2-phase model: Fluid A (e.g. air) d 2 3Ca B f ( , R ) cos h interface ds 2 h(h 3 ) θ Control parameters: Fluid B (e.g. water) Ca B U B R A B o :static contact angle (wettability) U 2 sin 3 [ R 2 ( 2 sin 2 ) 2 R{ ( ) sin 2 } {( ) 2 sin 2 }] f ( , R ) 3[ R (sin 2 2 ){( ) sin cos } {sin 2 ( ) 2 }( sin cos )] 2-phase model: Fluid A (e.g. air) d 2 3Ca B f ( , R ) cos h interface ds 2 h(h 3 ) θ Control parameters: Fluid B (e.g. water) Ca B U B R A B o :static contact angle (wettability) U Boundary conditions: 1. h (at the contact line) = 0 2. θ (at the contact line) = θo We use shooting method 3. θ (at the bath) = π/2 to find the solutions 2-phase model: Fluid A (e.g. air) d 2 3Ca B f ( , R ) cos h interface ds 2 h(h 3 ) θ Control parameters: Fluid B (e.g. water) Ca B U B R A B o :static contact angle (wettability) U Question: How CaBc depends on R and θo ? Part three: Results Control parameters: How is critical CaBc found? U B Ca B R A B e.g. fixed θo =50o , fixed R =0.1 o :static contact angle (wettability) 1 Static profile θo =50o 0.5 air 0 Δ -0.5 liquid -1 -1.5 -2 -2.5 0 0.05 0.1 0.15 0.2 0.25 Ca B Control parameters: How is critical CaBc found? U B Ca B R A B e.g. fixed θo =50o , fixed R =0.1 o :static contact angle (wettability) 1 0.5 air 0 Δ -0.5 liquid -1 -1.5 Uniform speed U -2 -2.5 0 0.05 0.1 0.15 0.2 0.25 Ca B Control parameters: How is critical CaBc found? U B Ca B R A B e.g. fixed θo =50o , fixed R =0.1 o :static contact angle (wettability) 1 0.5 air 0 -0.5 Δ liquid -1 -1.5 Uniform speed U -2 -2.5 0 0.05 0.1 0.15 0.2 0.25 Ca B Control parameters: How is critical CaBc found? U B Ca B R A B e.g. fixed θo =50o , fixed R =0.1 o :static contact angle (wettability) 1 0.5 air 0 -0.5 Δ liquid -1 -1.5 -2 Uniform speed U -2.5 0 0.05 0.1 0.15 0.2 0.25 Ca B Control parameters: How is critical CaBc found? U B Ca B R A B e.g. fixed θo =50o , fixed R =0.1 o :static contact angle (wettability) 1 0.5 air 0 -0.5 Δ liquid -1 -1.5 -2 Uniform speed U -2.5 0 0.05 0.1 0.15 0.2 0.25 Ca B Cac How does CaBc depend on R ? Control parameters: U B Ca B R A Critical capillary no. (Cac) B o :static contact angle (wettability) 1 R=1 0 R=0.1 R=0.01 fixed θo =50o -1 R=0.001 R=0 -2 -3 -4 -5 0 0.5 1 1.5 2 2.5 3 Ca Fluid A How does CaBc depend on R ? Fluid B 1 0 U -1 (fixed θo =50o) Log(Ca ) Bc -2 R A -3 B U B Ca B -4 -5 -4 -3 -2 -1 0 1 2 3 Log(R) Fluid A How does CaBc depend on R ? Fluid B 1 0 U -1 (fixed θo =50o) Log(Ca ) Bc -2 R A Dewetting regime B (-1 scaling) -3 U B Ca B -4 -5 -4 -3 -2 -1 0 1 2 3 Log(R) Fluid A How does CaBc depend on R ? Fluid B 1 0 U -1 Wetting regime (fixed θo =50o) Log(Ca ) Bc -2 R A -3 B U B Ca B -4 -5 -4 -3 -2 -1 0 1 2 3 Log(R) CaBc changes significantly with R, even for small air viscosity ! Fluid A How does CaBc depend on R ? Fluid B 1 What is the scaling ? 0 U -1 Wetting regime (fixed θo =50o) Log(Ca ) Bc -2 R A -3 B U B Ca B -4 -5 -4 -3 -2 -1 0 1 2 3 Log(R) CaBc changes significantly with R, even for small air viscosity ! Fluid A How does CaBc depend on R ? Fluid B 1 0 U -1 Wetting regime (fixed θo =50o) Log(Ca ) Bc -2 R A -3 B U B Ca B -4 -5 -4 -3 -2 -1 0 1 2 3 Log(R) Special case : R = 0 (i.e. log(R) → -infinity) How does CaBc depend on R ? Special case : R = 0 (i.e. log(R) → -infinity) d 2 3Ca B 2 2 f ( , R 0) cos ds h How does CaBc depend on R ? Special case : R = 0 (i.e. log(R) → -infinity) d 2 3Ca B 2 2 f ( , R 0) cos ds h Outer region (balance between gravity and viscous force) Asymptotic solution when CaB very large 3CaB cos 2 f ( ,0) as2 h How does CaBc depend on R ? Special case : R = 0 (i.e. log(R) → -infinity) d 2 3Ca B 2 2 f ( , R 0) cos ds h Outer region (balance between gravity and viscous force) Asymptotic solution when CaB very large 3CaB cos 2 f ( ,0) as2 h Inner region (balance between surface tension and viscous force) Asymptotic solution when CaB very large d 2 3Ca B 2 2 f ( ,0) b / sinner ds h How does CaBc depend on R ? inner inner Special case : R = 0 (i.e. log(R) → -infinity) d 2 3Ca B 2 2 f ( , R 0) cos ds h Outer region (balance between gravity and viscous force) Asymptotic solution when CaB very large 3CaB cos 2 f ( ,0) as2 h Inner region (balance between surface tension and viscous force) Asymptotic solution when CaB very large d 2 3Ca B 2 2 f ( ,0) b / sinner ds h Matching between inner region and outer region is always possible! How does CaBc depend on θo (wettability)? (fixed R = 0.01) 0.7 0.6 0.5 Cac Bc 0.4 Ca 0.3 0.2 0.1 0 0 0.5 1 1.5 2 2.5 3 o Critical speed decreases significantly for hydrophobic surface ! How does CaBc depend on θo (wettability)? (fixed R = 0.01) 0.7 0.6 0.5 Cac Bc 0.4 Ca 0.3 0.2 0.1 0 0 0.5 1 1.5 2 2.5 3 o Critical speed decreases significantly for hydrophobic surface ! (consistent with Duez et al. Nature Physics) Conclusion: 1. We developed a “lubrication-like” model for two- phase flow. 2. Air dynamics is crucial to find entrainment threshold. If air flow is neglected (i.e. R=0), there is no air entrainment no matter how large Ca is. 3. Asymptotic scaling of CaBc for small R? 1 ? 0 Dewetting -1 regime Log(Ca ) Bc -2 (-1 scaling) -3 -4 -5 -4 -3 -2 -1 0 1 2 3 Log(R) Conclusion: 1. We developed a “lubrication-like” model for two- phase flow. 2. Air dynamics is crucial to find entrainment threshold. If air flow is neglected (i.e. R=0), there is no air entrainment no matter how large Ca is. 3. Asymptotic scaling of CaBc for small R? Funded by: 1 ? 0 Dewetting -1 regime Log(Ca ) Bc Thank you! -2 (-1 scaling) -3 -4 -5 -4 -3 -2 -1 0 1 2 3 Log(R) (R)

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