Behavior of a Droplet on a Thin Liquid Film Ellen R. Peterson - PowerPoint PPT Presentation

Behavior of a Droplet on a Thin Liquid Film Ellen R. Peterson Carnegie Mellon University Department of Mathematical Sciences Center for Nonlinear Analysis February 23, 2012 Funded by National Science Foundation Ellen R. Peterson (Carnegie Mellon) Spreading Droplets February 23, 2012 1 / 35

Introduction Project Motivation Goal: Improve treatment for Cystic Fibrosis - use surfactant to spread medicine over mucus in lungs Research Group: CMU Physics: Steve Garoff, Fan Gao CMU Chemical Engineering: Todd Przybycien, Robert Tilton, Roomi Kalita, Ramankur Sharma, Amsul Khanal U. Pitt Medical Center: Tim Corcoran Questions: • How does the fluid spread? • Where is the surfactant? • How do multiple droplets interact? Ellen R. Peterson (Carnegie Mellon) Spreading Droplets February 23, 2012 2 / 35

Introduction Background on Spreading H D <<1 D z fluid H r substrate Drop spreading on a solid substrate: Multivalued velocity profile-precursor layer: Dussan, Davis (1974) Bertozzi (1998) Glasner (2003) Coarsening of droplets: Otto, Rump, Slepcev (2005) Witelski, Glasner (2003) Ellen R. Peterson (Carnegie Mellon) Spreading Droplets February 23, 2012 3 / 35

Surfactant Background on Spreading Surfactant Surfactant spreading on liquid subphase: Model: Gaver and Grotberg (1990) � 1 � 1 � 1 3 rh 3 � 1 h t + 1 2 rh 2 σ (Γ) r r = β 1 3 rh 3 h r r − κ 1 � � � � r ( rh r ) r r r r r r � 1 � 1 � 1 Γ t + 1 r ( rh Γ σ (Γ) r ) r = β 1 2 rh 2 Γ h r r − κ 1 2 rh 2 Γ r + δ 1 � � � r ( rh r ) r r ( r Γ r ) r r r r Asymptotic spreading behavior/similarity solutions: Jensen, Grotberg (1992), Jensen (1994) h t − 1 Γ t − 1 � 1 2 rh 2 Γ r � r = 0 r ( rh ΓΓ r ) r = 0 r Similarity scaling: Γ( r , t ) = 1 ξ = r h ( r , t ) = H ( ξ ) , G ( ξ ) , 1 1 t t 2 4 Similarity solutions: G ( ξ ) = − 1 ξ = r H ( ξ ) = 2 ξ 2 , 8 log ξ, 1 t 4 Ellen R. Peterson (Carnegie Mellon) Spreading Droplets February 23, 2012 4 / 35

Surfactant Numerical similarity solutions: ERP, Shearer (2012) Solutions: Equations: Γ( r , t ) = 1 r h t − ξ ˙ r 0 ( t ) 1 h ( r , t ) = H ( ξ ) , G ( ξ ) , ξ = � � 2 h 2 Γ ξ ξ 1 r 0 ( t ) h ξ = 1 r 0 ( t ) t r 0 ( t ) 2 2 ξ Γ t − ξ ˙ r 0 ( t ) 1 r 0 ( t ) Γ ξ = r 0 ( t ) 2 ( ξ h ΓΓ ξ ) ξ G ( ξ ) = − 1 ξ = r H ( ξ ) = 2 ξ 2 , 8 log ξ, 1 t − 1 4 r 0 ˙ = r 0 h (1 , t )Γ ξ (1 , t ) r ξ = r 0 ( t ) 2 4 1.5 3 H G h 1 Γ 2 0.5 1 0 0 0 0.2 0.4 0.6 0.8 1 1.2 0 0.2 0.4 0.6 0.8 1 1.2 r/r 0 (t) ξ r/r 0 (t) ξ 1000<t<10000 Ellen R. Peterson (Carnegie Mellon) Spreading Droplets February 23, 2012 5 / 35

Surfactant Inner Solution: ERP, Shearer (2011) � r � r � � 2 � 2 � 1 h ( r , t ) = t − 2 + O ( r 4 ) H 0 + + A 9 1 1 18 G 0 t t 6 9 � r � � 2 � G 0 + µ Γ( r , t ) = t − 4 + O ( r 4 ) 9 1 H 0 t 6 0.12 0.65 0.1 t 4/9 Γ ( η ,t) 0.08 h( η ,t) 0.6 0.06 0.04 0.02 0.55 0 0.2 0.4 0.6 0 0.1 0.2 0.3 0.4 η η Ellen R. Peterson (Carnegie Mellon) Spreading Droplets February 23, 2012 6 / 35

Surfactant Experimental Goals Experimental spreading: Bull et al (1999), Fallest et al (2010) Visualize the height of the film (using laser line) Visualize the location of the surfactant (fluorescence) Match the experimental data to the PDE model ring 2 4 2 4 (a) t = 1 sec (b) t = 2 sec Γ [ μ g/cm 2 ] Γ [ μ g/cm 2 ] H [mm] H [mm] 1 2 1 2 0 0 0 0 0 0 20 20 40 40 0 0 20 20 40 40 r [mm] r [mm] 2 4 2 4 (c) t = 3 sec (d) t = 6 sec Γ [ μ g/cm 2 ] Γ [ μ g/cm 2 ] H [mm] H [mm] 1 2 1 2 0 0 0 0 0 0 20 20 40 40 0 0 20 20 40 40 r [mm] r [mm] Karen Daniels’s Lab NC State University Ellen R. Peterson (Carnegie Mellon) Spreading Droplets February 23, 2012 7 / 35

Surfactant Spreading Behavior tracking tracking 2 4 2 4 (b) t = 2 sec (b) t = 2 sec Γ [ μ g/cm 2 ] Γ [ μ g/cm 2 ] H [mm] H [mm] 1 2 1 2 0 0 0 0 0 0 20 20 40 40 0 0 20 20 40 40 r [mm] r [mm] 10 2 2 10 λ=0.25 λ=0.295 r M [mm] r 0 [mm] 1 10 1 10 −1 0 1 0 1 10 10 10 10 10 t [sec] t [sec] Ellen R. Peterson (Carnegie Mellon) Spreading Droplets February 23, 2012 8 / 35

Surfactant Surfactant Initial Condition Using time scale t = t ∗ : 1 1 0.6 0.8 Surfactant at 0.6 0.4 Γ Γ Γ t=9 0.5 0.4 0.2 0.2 0 0 0 2 4 6 0 0 2 4 6 0 2 4 6 r r r Initial Conditions 2 2 2 1.5 1.5 1.5 1.5 1 h Γ 1 Γ 1 Γ 1 0.5 0.5 0.5 0.5 0 0 0 0 2 4 6 0 2 4 6 0 2 4 6 0 2 4 6 r r r r 1.5 1.5 1.5 Height at 1 h 1 h h 1 t=9 0.5 0.5 0.5 0 2 4 6 0 2 4 6 0 2 4 6 r r r Ellen R. Peterson (Carnegie Mellon) Spreading Droplets February 23, 2012 9 / 35

Spreading Droplet How do we predict the spreading behavior? Spreading Parameter: (Harkins 1952) D Σ F Σ S = Σ F − Σ DF − Σ D DF Σ • Complete spreading ( S > 0) - Spreading on a Deep Layer: Dipietro, Huh, & Cox, 1978; DiPietro & Cox, 1980; Foda & Cox, 1980 S>0 - Spreading Power Law: Fraaije & Cazabat, 1989 • Lens-shape ( S < 0) - Numerical axisymmetric solution, S<0 static lens : Pujado & Scriven, 1971 Ellen R. Peterson (Carnegie Mellon) Spreading Droplets February 23, 2012 10 / 35

Spreading Droplet Assumptions • Newtonian drop on Newtonian fluid • Fluids immiscible • Lubrication Approximation H D << 1 • S < 0 • axisymmetric spreading Contact Line Force Balance Issue: drop r D 0 z fluid H r substrate Ellen R. Peterson (Carnegie Mellon) Spreading Droplets February 23, 2012 11 / 35

Spreading Droplet Deriving Droplet Spreading Equations r = µ F u F p D r = µ D u D p DF = µ DF u DF p F zz , zz , r zz p D z = − ρ D g , p DF = − ρ DF g , p F z = − ρ F g z p DF ( b , t ) = p D ( b , t ) − Σ DF b rr , p F ( h , t ) = p atm − Σ F h rr p D ( a , t ) = p atm − Σ D a rr , a(r,t) D u =0 z u =0 F h(r,t) z DF DF D D μ u =μ u z z DF u =u D b(r,t) DF F u =0 u =0 Ellen R. Peterson (Carnegie Mellon) Spreading Droplets February 23, 2012 12 / 35

Spreading Droplet Droplet Spreading Equations - axisymmetric 1 � � � 1 � � � h 3 h t = r Boh r − r ( rh r ) r r r r 1 � � � � 1 � � b 3 Bo (1 − β ) b r − σ DF b t = r r ( rb r ) r r r � b 2 � � 1 � �� Bo β a r − σ D + r ( ra r ) r 2 (3 a − b ) r r � b 2 � �� � 1 � 1 Bo (1 − β ) b r − σ DF = r ( rb r ) r 2 (3 a − b ) a t r r r � 1 − λ � � � 1 � ��� ( a − b ) 3 + a 3 Bo β a r − σ D + r ( ra r ) r λ r r Bo: Bond number β : density ratio λ : viscosity ratio σ D : surface tension of drop and air σ DF : surface tension of drop and fluid Ellen R. Peterson (Carnegie Mellon) Spreading Droplets February 23, 2012 13 / 35 related work: Kriegsmann & Miksis 2003

Equilibrium Solutions Equilibrium Solutions � 1 � r ( rh r ) r = 0 Boh r − r � 1 � Bo β a r − σ D r ( ra r ) r = 0 r � 1 � Bo (1 − β ) b r − σ DF r ( rb r ) r = 0 . r If β � = 1 these equations are Bessel equations. We’ll assume β < 1 � √ � √ � � h ( r ) = C h + C hh I 0 Bor + C hhh K 0 Bor �� � �� � Bo β Bo β a ( r ) = C a + C a I 0 σ D r + C aaa K 0 σ D r �� � �� � Bo (1 − β ) Bo (1 − β ) b ( r ) = C b + C bb I 0 r + C bbb K 0 r σ DF σ DF Related studies: Kriegsmann & Miksis (2003), Pujado & Scriven (1971) Ellen R. Peterson (Carnegie Mellon) Spreading Droplets February 23, 2012 14 / 35

Equilibrium Solutions Boundary Conditions Height of Film: Edge Conditions: � r 0 � L a r (0 , t ) = 0 , b r (0 , t ) = 0 , h r ( L , t ) = 0 2 π b ( r , t ) rdr + 2 π h ( r , t ) rdr = Mass F 0 r 0 Continuity of Interface: Location of r 0 ( t ): a ( r 0 , t ) = h ( r 0 , t ) , b ( r 0 , t ) = h ( r 0 , t ) � r 0 p DF = p F | r = r 0 Mass D = 2 π ( a ( r ) − b ( r )) r dr � � Continuity of Pressure: 0 � ∆ h − σ D ∆ a − σ DF ∆ b = 0 � � r = r 0 a(r,t) r (t) 0 h(r,t) b(r,t) Ellen R. Peterson (Carnegie Mellon) Spreading Droplets February 23, 2012 15 / 35

Equilibrium Solutions Boundary Conditions Balance of Surface Tension Forces: Σ F cos θ F + Σ D cos θ D + Σ DF cos θ DF = 0 Σ F sin θ F + Σ D sin θ D + Σ DF sin θ DF = 0 Apply approximations: θ F ≈ ε h r , θ D ≈ − ε a r , θ DF ≈ − ε b r 1 + ε 2 h 2 1 + ε 2 a 2 1 + ε 2 b 2 � � � � � � Σ F r − Σ D r − Σ DF r + . . . + . . . + . . . = 0 2 2 2 ε h r + ε 3 h 3 ε a r + ε 3 a 3 ε b r + ε 3 b 3 � � � � � � Σ F r − Σ D r − Σ DF r + . . . + . . . + . . . = 0 3 3 3 D Σ O (1) : 1 − σ D − σ DF = 0 D θ θ F ⇒ S ∼ O ( ε 2 ) = F Σ O ( ε ) : h r − σ D a r − σ DF b r = 0 2π-θ DF O ( ε 2 ) : h 2 r − σ D a 2 r − σ DF b 2 r = − 2 DF Σ Ellen R. Peterson (Carnegie Mellon) Spreading Droplets February 23, 2012 16 / 35