Bretherton s Problem. Flow of long bubbles in tubes. Prof. Michiel T. Kreutzer Delft University of Technology


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1 Bretherton s Problem Flow of long bubbles in tubes Prof. Michiel T. Kreutzer Delft University of Technology
2 Every drop/bubble is a test tube Distribute Optical valve Splitup Cristobal et al., App. Phys. Lett. (006) Baroud et al., Lab Chip (007) Synchronize Sort Link et al., Phys. Rev. Lett. (004) Merge Feedback Link et al., Ang. Chem. Int. Ed. (006) Prakash et al., Science (007) Baroud et al., Lab Chip (007) Prakash et al., Science (007)
3 Labonachip Chemistry Ismagilov, 006 Kreutzer et al., Chem Eng Sci 60 (005) 5895 Gunther et al., Lab Chip 6 (006) 1487
4 4
5 5
6 Our basic questions for today How large are the bubbles/drops? > Next class, on bubble/droplet formation What is the flow resistance? Film thickness? Inertia, Marangoni effects? Partial wetting? Channel Shape? 6
7 Bretherton = LandauLevich y x u Moving reference frame: bubble stands still: u(0) = U, ( h) = 0 y Navier Stokes: Continuity p u = μ x r h q + = 0, with q = udy t x h 0 7
8 Bretherton = LandauLevich y x Navier Stokes: y yh u= U+ p μ μ Laplace pressure p = [ γh ] Continuity γ t x μ x x h h h + U h = 0 h γ h q= udy = Uh h μ x 0 8
9 Bretherton = LandauLevich γ t x μ x x h h h + U h = 0 y x Bubble stands still, so must be zero, we balance the nd and rd terms: h γ h ~ h x μu x x x ~ λ δ γ 1 δ 1/ δ ~ δ or Ca ~ h~ δ λ μu λ λ λ 9
10 Bretherton = LandauLevich First matching rule 1/ ~ δ Ca λ 1 κ = r y x Scaling of the problem δ ~ rca λ ~ rca / 1/ h δ κ ~ ~ 1 κ = x λ r Second matching rule δ 1 ~ r λ 10
11 Full Solution of the simplest case: LandauLevichBretherton film equation
12 Scaling of the film equation We start from the film evolution equation h γ h t x μ x + Uh h = 0 y x Steady state, so drop the transient term γ μ h x Uh h = c h μu h c = x γ h In the flat region, h 0 = x h η = δ ( Ca) ξ = δ 1/ x η η 1 = η which gives c: δ c 0 =, δ c= δ 1
13 Forced Wetting Film Equation exponential range η 1 η η 1 = η parabolic range η 1 η η 1 which has an analytical solution η 0 which has an analytical solution 1/ 1/ ξ ξ ξ η = 1+ αe + βe cos ξ + γe sin ξ P η = ξ + Qξ+ R 1
14 Four different regions with their scaling η η 1 = η y x flat exponential parabolic spherical no η = 1 show n η 0 n = no η 1 show η η 1 η η 1, 1 r/ δ η 1/ η Ca ( ), 0 η ~1 r/ δ η r( Ca) δ = / 14
15 Set up match of exp and parabolic η η 1 = η Exponential: η η 1 = η( ξ) = αe ξ y x parabolic: η 0 P ( ) = + + η ξ ξ Qξ R flat exponential parabolic spherical no η = 1 show n η 0 n = no η 1 show η η 1 η η 1, 1 r/ δ η 1/ η Ca ( ), 0 η ~1 r/ δ η r( Ca) δ = / 15
16 Use exponential for initial conditions: η( ξ) = 1+ ae η(0) = 1+ a η (0) = a η (0) = a η η 1 = η ξ And integrate full equation with that Numerical Shooting 0.64 ηξ = ξ + ( ) e ξ Numerical solution Move numerical solution to left and right, such that Q = 0 Fitting: P = 0.64, R =.79 Note: MTK gets a = 0.001, P = 0.64, Q=0, R =.90 with Mathematica s NDSolve P ( } = + + P= η ( ξ) ηξ ξ Qξ R Q = η ( ξ) ξη ( ξ) R = η ξ ξ + ξ η ξ η ξ ( ) ( ) ( ) ( )
17 Set up match of parabolic and spherical η η 1 = η h = x 1 r η δ = / r( Ca) is the same as y x Parabola matches to the sphere: δ δ 0.64= = 0.64( Ca) / r( Ca) r / flat exponential parabolic spherical no η = 1 show n η 0 n = η = 1+ αe ξ η= ξ + η = a n η ~1 r/ δ η r( Ca) δ = / 17
18 Pressure Jump x / The parabola h= ( Ca) r matches to a sphere of curvature ( / r)( ( Ca) / y x flat exponential parabolic spherical no η = 1 show n η 0 n = η = 1+ αe ξ η= ξ + η = a n η ~1 r/ δ η r( Ca) δ = / 18
19 Rear of the bubble ηξ = ξ ( ) Numerical solution 44 Start numerical solution with exponential function 1/ 1/ ξ ξ η = 1+ βe cos ξ + γe sin ξ  19
20 Pressure drop over the bubble: / Δ p= [ ( Ca) ] γ r / Δ p= [1 0.46( Ca) ] γ r Δp= 4.5( Ca) / γ r 0
21 Experiments / δ= 0.64( Ca) r Taylor, J Fluid Mech 10 (1961), 161 Bretherton, J Fluid Mech 10 (1961), 166 1
22 The notsosimple cases Marangoni, Inertia, nonround, thick films,
23 Why the maximum? First matching rule 1/ ~ δ Ca λ Second matching rule δ ~ 1 λ r δ Thick film decreases radius of curvature! Scaling of the problem / Ca δ ~ r 1 + Ca / / Ca λ~ r [1 + Ca ] / 1/ Aussilous, Phys Fluids 1 (000) 67 Taylor, J Fluid Mech 10 (1961), 161
24 Marangoni Effects y x u Moving reference frame: bubble stands still: u(0) = U, ( h) = 0 y u(0) = U, u( h) = 0 now h γ h q= udy = Uh h 1μ x 0 4
25 Marangoni Effects y x u Moving reference frame: bubble stands still: u(0) = U, ( h) = 0 y 5 γ t x μ x x h h h + U h = 0 γ 1μ u(0) = U, u( h) = 0 / δ= 0.64 r( Ca) / δ= 0.64 r(1 Ca)
26 Inertia changes bubble shape Ca = 0.04 Re = 1,10,100,00 Kreutzer et al, AIChE J 51 (005) 48 6
27 Inertia increases film thickness δ ~ r 1 We = Ca Ca / / + ( ) γ ρu r δ We Kreutzer et al, AIChE J 51 (005) 48 Han and Shikazono, Int J. Heat Fluid Flow (009) Aussilous, Phys Fluids 1 (000) 67 7
28 Square Channels 8
29 Hydrostatics Without flow, there cannot be a gradient of (Laplace) pressure p = 0 For a channel of width W and height H, we have at the nose Δ p= γ + H W 1 In the gutter, no curvature in the axial direction, so 1 1 Δ p= γ + α 1 α = + W H 9 Ajaev & Homsy, Ann Rev. Fluid Mech 8 (006) 77 Wong, Radke & Morris, J Fluid Mech 9 (1995) 71
30 Film Deposition 0 Wong, Radke & Morris, J Fluid Mech 9 (1995) 71
31 The nonuniformity starts at the nose 0.64 δ (Ca cos φ) / = α 1 Wong, Radke & Morris, J Fluid Mech 9 (1995) 71
32 What have we learned Scaling in flow of long bubbles / droplets in cylindrical tubes (basically the same as Landau Levich) Full solution requires a double matching Marangoni effects increase film thickness and pressure drop by a factor 4 / There is a maximum film thickness at high Ca (this does not happen with a flat plate LL scaling) Inertia increases film thickness, numerical solution needed Square channels have highly nonuniform films
33 Further Reading Recent reviews Ajaev & Homsy, Ann Rev. Fluid Mech 8 (006) 77 Kreutzer et al., Chem Eng Sci 60 (005) 5895 Gunther et al., Lab Chip 6 (006) 1487 Microfluidics: a couple of key papers Wong, Radke & Morris, J Fluid Mech 9 (1995) 71 Aussilous, Phys Fluids 1 (000) 67
34 Extra: partial wetting in microchannels
35 5
36 Controlled experiments to study instability 6
37 Constant dewetting velocity γ 1 Vd = kθe k= μ 1 ln( r/ a) 7
38 Film rupture γ t x μ x 6πh x h h A h + h + = 0 8 Oron et al, Rev Mod Phys 69 (1997) 91
39 Plateau Borders 9 Koehler et al, Phys Rev Lett 8 (1999) 4 Koehler et al, Phys Rev E 66 (00) (R)
40 In the blink of an eye ~ 40 s h γ h t x 1μ x + h = 0 40
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